#set document(title: "3.5 Irrational Numbers", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 3.5#h(0.6em)Irrational Numbers #figure(figph[A painting shows Pythagoreans' hymn to the rising sun.], alt: "A painting shows Pythagoreans' hymn to the rising sun.", caption: [The Pythagoreans were a philosophical sect of ancient Greece, often associated with mathematics.]) === Learning Objectives After completing this section, you should be able to: + Define and identify numbers that are irrational. + Simplify irrational numbers and express in lowest terms. + Add and subtract irrational numbers. + Multiply and divide irrational numbers. + Rationalize fractions with irrational denominators. The Pythagoreans were a philosophical sect in ancient Greece. Their philosophy included reincarnation and purifying the mind through the study and contemplation of mathematics and science. One of their principles was the cosmos is ruled by order, specifically mathematics and music. They even held mystic beliefs about specific numbers and figures. For example, the number 1 was associated with the mind and essence. Four represented justice, as it is the first product of two even numbers. Most famously, though, is the association with the Pythagorean Theorem, which states that in a right triangle, the sum of the squares of the shorter sides of the triangle (the legs) equals the square of the longer side (the hypotenuse). Even the ancient Egyptians used this relationship, as triangles with side measures 3, 4, and 5 were often used in surveying following the flooding of the Nile. There is a story of a Pythagorean, Hippasus, discovering that not all numbers could be expressed as fractions. In other words, not all numbers were rational numbers. The story ends with Hippasus, who shared this, or in some versions discovered it, put to death by drowning for sharing this fact, that not all quantities could be expressed as the ratio of two natural numbers. As colorful as that story may be, it is most likely false, as there are no contemporary sources to corroborate it. But it does seem to mark the discovery that not all quantities or measures were fractions of numbers. And so, #strong[irrational numbers] were discovered. #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Philosophy_of_Pythagoreans")[The Philosophy of the Pythagoreans] ] === Defining and Identifying Numbers That Are Irrational We defined rational numbers in the last section as numbers that could be expressed as a fraction of two integers. #strong[Irrational numbers] are numbers that cannot be expressed as a fraction of two integers. Recall that rational numbers could be identified as those whose decimal representations either terminated (ended) or had a repeating pattern at some point. So irrational numbers must be those whose decimal representations do not terminate or become a repeating pattern. One collection of irrational numbers is #strong[square roots] of numbers that aren’t #strong[perfect squares]. #math.equation(block: false, alt: "x")[$x$] is the square root of the number #math.equation(block: false, alt: "a")[$a$], denoted #math.equation(block: false, alt: "the square root of a")[$sqrt(a)$], if #math.equation(block: false, alt: "x squared equals a")[$x^(2) = a$]. The number #math.equation(block: false, alt: "a")[$a$] is the perfect square of the integer #math.equation(block: false, alt: "n")[$n$] if #math.equation(block: false, alt: "a equals n squared")[$a = n^(2)$]. The rational number #math.equation(block: false, alt: "the fraction a over b")[$frac(a, b)$] is a perfect square if both #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are perfect squares. One method of determining if an integer is a perfect square is to examine its prime factorization. If, in that factorization, all the prime factors are raised to even powers, the integer is a perfect square. Another method is to attempt to factor the integer into an integer squared. It is possible that you recognize the number as a perfect square (such as 4 or 9). Or, if you have a calculator at hand, use the calculator to determine if the square root of the integer is an integer. #examplebox("Example 1")[Identifying Perfect Squares][ Determine which of the following are perfect squares. + 45 + 81 + #math.equation(block: false, alt: "the fraction 9 over 28")[$frac(9, 28)$] + #math.equation(block: false, alt: "the fraction 144 over 400")[$frac(144, 400)$] #solutionbox[ + The prime factorization of 45 is #math.equation(block: false, alt: "45 equals 3 squared times 5")[$45 = 3^(2) × 5$]. Since the 5 is not raised to an even power, 45 is not a perfect square. + The prime factorization of 81 is #math.equation(block: false, alt: "3 to the power 4")[$3^(4)$]. All the prime factors are raised to even powers, so 81 is a perfect square. + We must determine if both the numerator and denominator of #math.equation(block: false, alt: "the fraction 9 over 28")[$frac(9, 28)$] are perfect squares for the rational number to be a perfect square. The numerator is 9, and as mentioned above, 9 is a perfect square (it is 3 squared). Now we check the prime factorization of the denominator, 28, which is #math.equation(block: false, alt: "28 equals 2 squared times 7")[$28 = 2^(2) × 7$]. Since 7 is not raised to an even power, 28 is not a perfect square. Since the denominator is not a perfect square, #math.equation(block: false, alt: "the fraction 9 over 28")[$frac(9, 28)$] is not a perfect square. + We must determine if both the numerator and denominator of #math.equation(block: false, alt: "the fraction 144 over 400")[$frac(144, 400)$] are perfect squares for the rational number to be a perfect square. The numerator is 144. The prime factorization of 144 is #math.equation(block: false, alt: "144 equals 2 to the power 4 times 3 squared")[$144 = 2^(4) × 3^(2)$]. Since all the prime factors of 144 are raised to even powers, 144 is a perfect square. Now we check the prime factorization of the denominator, 400, which is #math.equation(block: false, alt: "400 equals 2 to the power 4 times 5 squared")[$400 = 2^(4) × 5^(2)$]. Since all the prime factors of 400 are raised to even powers, 400 is a perfect square. Since the numerator and denominator of #math.equation(block: false, alt: "the fraction 144 over 400")[$frac(144, 400)$] are perfect squares, #math.equation(block: false, alt: "the fraction 144 over 400")[$frac(144, 400)$] is a perfect square. ] ] #notebox("Tech Check", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[ #emph[Using Desmos to Determine if a Number Is a Perfect Square] Desmos may be used to determine if a number is a perfect square by using its square root function. When Desmos is opened, there is a tab in the lower left-hand corner of the Desmos screen. This tab opens the Desmos keypad, shown. #figure(figph[Desmos keyboard is displayed. Three sets of keys are displayed. The first set has 4 rows of 4 keys, each. Row 1: x, y, a squared, and a to the power b. Row 2: open parenthesis, close parenthesis, lesser than, and greater than. Row 3: modulus of a, comma, lesser than or equal to, and greater than or equal to. Row 4: A B C, sound, square root, which is circled, and pi. The second set has 4 rows of 4 keys, each. Row 1: 7, 8, 9, and division symbol. Row 2: 4, 5, 6, and multiplication symbol. Row 3: 1, 2, 3, and minus symbol. Row 4: 0, dot, equals sign, and plus sign. The third set has the following keys: functions, left arrow, right arrow, close, and back.], alt: "Desmos keyboard is displayed. Three sets of keys are displayed. The first set has 4 rows of 4 keys, each. Row 1: x, y, a squared, and a to the power b. Row 2: open parenthesis, close parenthesis, lesser than, and greater than. Row 3: modulus of a, comma, lesser than or equal to, and greater than or equal to. Row 4: A B C, sound, square root, which is circled, and pi. The second set has 4 rows of 4 keys, each. Row 1: 7, 8, 9, and division symbol. Row 2: 4, 5, 6, and multiplication symbol. Row 3: 1, 2, 3, and minus symbol. Row 4: 0, dot, equals sign, and plus sign. The third set has the following keys: functions, left arrow, right arrow, close, and back.", caption: [Desmos keyboard with square root key circles]) There you find the key for the square root, which is circled in . To find the square root of a number, click the square root key, which begins a calculation, and then enter the value for which you want a square root. If the result is an integer, then the number is a perfect square. ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/square_root_of_a_number")[Using Desmos to Find the Square Root of a Number] ] Another collection of irrational numbers is based on the special number, #strong[pi], denoted by the Greek letter #math.equation(block: false, alt: "π")[$π$], which is the ratio of the circumference of the diameter of the circle . #figure( stdfig([#cetz.canvas({ import cetz.draw: * circle((100.00pt, 100.00pt), radius: 60.00pt, fill: none, stroke: 0.7pt + black) line((40.00pt, 100.00pt), (160.00pt, 100.00pt), stroke: (paint: black, thickness: 0.6pt, dash: "dashed")) line((100.00pt, 100.00pt), (160.00pt, 100.00pt), stroke: 0.7pt + black) circle((100.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none) content((104.00pt, 104.00pt), anchor: "south-west", text(size: 9pt)[O]) circle((160.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none) content((164.00pt, 100.00pt), anchor: "west", text(size: 9pt)[A]) circle((40.00pt, 100.00pt), radius: 1.8pt, fill: black, stroke: none) content((36.00pt, 100.00pt), anchor: "east", text(size: 9pt)[B]) content((70.00pt, 112.00pt), anchor: "center", text(size: 9pt)[diameter]) content((130.00pt, 86.00pt), anchor: "center", text(size: 9pt)[radius]) content((100.00pt, 20.00pt), anchor: "center", text(size: 9pt)[Circumference]) })]), alt: "A circle with its radius, diameter, and circumference labeled.", caption: [Circle with radius, diameter, and circumference labeled], ) Any multiple or power of #math.equation(block: false, alt: "π")[$π$] is an irrational number. Any number expressed as a rational number times an irrational number is an irrational number also. When an irrational number takes that form, we call the rational number the #strong[rational part], and the irrational number the #strong[irrational part]. It should be noted that a rational number plus, minus, multiplied by, or divided by any irrational number is an irrational number. #examplebox("Example 2")[Identifying Irrational Numbers][ Identify which of the following numbers are irrational. + #math.equation(block: false, alt: "the square root of 35")[$sqrt(35)$] + #math.equation(block: false, alt: "0. 15 ¯")[$0. overline(15)$] + #math.equation(block: false, alt: "the square root of 121")[$sqrt(121)$] + #math.equation(block: false, alt: "4 π")[$4 π$] #solutionbox[ + 35 can be factored as #math.equation(block: false, alt: "5 times 7")[$5 × 7$], showing that 35 is not the square of an integer or a rational number. This mean its square root is an irrational number. + Since #math.equation(block: false, alt: "0. 15 ¯")[$0. overline(15)$] is a decimal with a repeating pattern, it is rational, so it is not an irrational number. + #math.equation(block: false, alt: "121 equals 11 squared")[$121 = 11^(2)$]. Since 121 is the square of an integer, its square root is a rational number. + Since #math.equation(block: false, alt: "4 π")[$4 π$] is a multiple of pi, it is irrational. In this case, the rational part of the number is 4, while the irrational part is #math.equation(block: false, alt: "π")[$π$]. ] ] #notebox("Who Knew?", rgb("#183B6F"), rgb("#183B6F"), rgb("#EFF1F5"))[ #emph[Euler-Mascheroni Constant] Determining if a number is rational or irrational is not trivial. There are numbers that defied such classification for quite a long time. One such is the Euler-Mascheroni constant. The Euler-Mascheroni constant is used in mathematics, and is primarily associated with the natural logarithm, which is a mathematical function. The constant has been around since around 1790. However, it was unknown if this constant was rational or irrational until 2013, at which point it was proven to be irrational. ] === Simplifying Square Roots and Expressing Them in Lowest Terms To #strong[simplify a square root] means that we rewrite the square root as a rational number times the square root of a number that has no perfect square factors. The act of changing a square root into such a form is simplifying the square root. The number inside the square root symbol is referred to as the #strong[radicand]. So in the expression #math.equation(block: false, alt: "the square root of a")[$sqrt(a)$] the number #math.equation(block: false, alt: "a")[$a$] is referred to as the radicand. Before discussing how to simplify a square root, we need to introduce a rule about square roots. The square root of a product of numbers equals the product of the square roots of those number. Written symbolically, #math.equation(block: false, alt: "the square root of a times b equals the square root of a times the square root of b")[$sqrt(a × b) = sqrt(a) × sqrt(b)$]. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ For any two numbers #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], #math.equation(block: false, alt: "the square root of a times b equals the square root of a times the square root of b")[$sqrt(a × b) = sqrt(a) × sqrt(b)$]. ] Using this formula, we can factor an integer inside a square root into a perfect square times another integer. Then the square root can be applied to the perfect square, leaving an integer times the square root of another integer. If the number remaining under the square root has no perfect square factors, then we’ve simplified the irrational number into lowest terms. To simplify the irrational number into lowest terms when #math.equation(block: false, alt: "n")[$n$] is an integer: #strong[Step 1:] Determine the largest perfect square factor of #math.equation(block: false, alt: "n")[$n$], which we denote #math.equation(block: false, alt: "a squared")[$a^(2)$]. #strong[Step 2:] Factor #math.equation(block: false, alt: "n")[$n$] into #math.equation(block: false, alt: "a squared times b")[$a^(2) × b$]. #strong[Step 3:] Apply #math.equation(block: false, alt: "the square root of a squared times b equals the square root of a squared times the square root of b")[$sqrt(a^(2) × b) = sqrt(a^(2)) × sqrt(b)$]. #strong[Step 4:] Write #math.equation(block: false, alt: "the square root of n")[$sqrt(n)$] in its simplified form, #math.equation(block: false, alt: "a the square root of b")[$a sqrt(b)$]. When a square root has been simplified in this manner, #math.equation(block: false, alt: "a")[$a$] is referred to as the rational part of the number, and #math.equation(block: false, alt: "the square root of b")[$sqrt(b)$] is referred to as the irrational part. #examplebox("Example 3")[Simplifying a Square Root][ Simplify the irrational number #math.equation(block: false, alt: "the square root of 180")[$sqrt(180)$] and express in lowest terms. Identify the rational and irrational parts. #solutionbox[ Begin by finding the largest perfect square that is a factor of 180. We can do this by writing out the factor pairs of 180: #math.equation(block: true, alt: "1 times 180 2 times 90 3 times 60 4 times 45 5 times 36 6 times 30 9 times 20 10 times 18 12 times 15")[$1 × 180 #h(1.5em) 2 × 90 #h(1.5em) 3 × 60 #h(1.5em) 4 × 45 #h(1.5em) 5 × 36 #h(1.5em) 6 × 30 #h(1.5em) 9 × 20 #h(1.5em) 10 × 18 #h(1.5em) 12 × 15$] Looking at the list of factors, the perfect squares are 4, 9, and 36. The largest is 36, so we factor the into #math.equation(block: false, alt: "36 times 5 equals 6 squared times 5")[$36 × 5 = 6^(2) × 5$]. In the formula, #math.equation(block: false, alt: "a equals 6")[$a = 6$] and #math.equation(block: false, alt: "b equals 5")[$b = 5$]. Apply #math.equation(block: false, alt: "the square root of a squared times b equals the square root of a squared times the square root of b")[$sqrt(a^(2) × b) = sqrt(a^(2)) × sqrt(b)$]. #math.equation(block: true, alt: "the square root of 6 squared times 5 equals the square root of 6 squared times the square root of 5")[$sqrt(6^(2) × 5) = sqrt(6^(2)) × sqrt(5)$] The simplified form of #math.equation(block: false, alt: "the square root of 180")[$sqrt(180)$] is #math.equation(block: false, alt: "6 the square root of 5")[$6 sqrt(5)$]. In this example, the 6 is the rational part, and the #math.equation(block: false, alt: "the square root of 5")[$sqrt(5)$] is the irrational part. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Simplifying_Square_Roots")[Simplifying Square Roots] ] #examplebox("Example 4")[Simplifying a Square Root][ Simplify the irrational number #math.equation(block: false, alt: "the square root of 330")[$sqrt(330)$] and express in lowest terms. Identify the rational and irrational parts. #solutionbox[ Begin by finding the largest perfect square that is a factor of 330. We can do this by writing out the factor pairs of 330: #math.equation(block: true, alt: "1 times 330 2 times 165 3 times 110 5 times 66 6 times 55 10 times 33 11 times 30 15 times 22")[$1 × 330 #h(1.5em) 2 × 165 #h(1.5em) 3 × 110 #h(1.5em) 5 × 66 #h(1.5em) 6 × 55 #h(1.5em) 10 × 33 #h(1.5em) 11 × 30 #h(1.5em) 15 × 22$] Looking at the list of factors, there are no perfect squares other than 1, which means #math.equation(block: false, alt: "the square root of 330")[$sqrt(330)$] is already expressed in lowest terms. In this case, 1 is the rational part, and #math.equation(block: false, alt: "the square root of 330")[$sqrt(330)$] is the irrational part. #strong[Though we could write this as #math.equation(block: false, alt: "1 the square root of 330")[$1 sqrt(330)$], but the product of 1 and any other number is just the number.] ] ] #examplebox("Example 5")[Simplifying a Square Root][ Simplify the irrational number #math.equation(block: false, alt: "the square root of 2,548")[$sqrt(2","548)$] and express in lowest terms. Identify the rational and irrational parts. #solutionbox[ Begin by finding the largest perfect square that is a factor of 2,548. We can do this by writing out the factor pairs of 2,548: #math.equation(block: true, alt: "1 times 2548 2 times 1274 4 times 637 7 times 364 13 times 196 14 times 182 26 times 98 28 times 91 49 times 52")[$1 × 2548 #h(1.5em) 2 × 1274 #h(1.5em) 4 × 637 #h(1.5em) 7 × 364 #h(1.5em) 13 × 196 #h(1.5em) 14 × 182 #h(1.5em) 26 × 98 #h(1.5em) 28 × 91 #h(1.5em) 49 × 52$] Looking at the list of factors, the perfect squares are 4, 49, and 196. The largest is 196, so we factor the 2,548 into #math.equation(block: false, alt: "196 times 13 equals 14 squared times 13")[$196 × 13 = 14^(2) × 13$]. In the formula, #math.equation(block: false, alt: "a equals 14")[$a = 14$] and #math.equation(block: false, alt: "b equals 5")[$b = 5$]. Apply #math.equation(block: false, alt: "the square root of a squared times b equals the square root of a squared times the square root of b")[$sqrt(a^(2) × b) = sqrt(a^(2)) × sqrt(b)$]. #math.equation(block: true, alt: "the square root of 14 squared times 13 equals the square root of 14 squared times the square root of 13")[$sqrt(14^(2) × 13) = sqrt(14^(2)) × sqrt(13)$] The simplified form of #math.equation(block: false, alt: "the square root of 2,548")[$sqrt(2","548)$] is #math.equation(block: false, alt: "14 the square root of 13")[$14 sqrt(13)$]. In this example, 14 is the rational part, and #math.equation(block: false, alt: "the square root of 13")[$sqrt(13)$] is the irrational part. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Simplifying_Square_Roots")[Simplifying Square Roots] ] === Adding and Subtracting Irrational Numbers Just like any other number we’ve worked with, irrational numbers can be added or subtracted. When working with a calculator, enter the operation and a decimal representation will be given. However, there are times when two irrational numbers may be added or subtracted without the calculator. This can happen only when the irrational parts of the irrational numbers are the same. To add or subtract two irrational numbers that have the same irrational part, add or subtract the rational parts of the numbers, and then multiply that by the common irrational part. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ Let our first irrational number be #math.equation(block: false, alt: "a times x")[$a × x$], where #math.equation(block: false, alt: "a")[$a$] is the rational and #math.equation(block: false, alt: "x")[$x$] the irrational parts. Let the other irrational number be #math.equation(block: false, alt: "b times x")[$b × x$], where #math.equation(block: false, alt: "b")[$b$] is the rational and #math.equation(block: false, alt: "x")[$x$] the irrational parts. Then #math.equation(block: false, alt: "a times x plus or minus b times x equals open parenthesis a plus or minus b close parenthesis times x")[$a × x ± b × x = ( a ± b ) × x$]. ] #examplebox("Example 6")[Subtracting Irrational Numbers with Similar Irrational Parts][ If possible, subtract the following irrational numbers without using a calculator. If this is not possible, state why. #math.equation(block: true, alt: "3 the square root of 7 minus 8 the square root of 7")[$3 sqrt(7) − 8 sqrt(7)$] #solutionbox[ Since these two irrational numbers have the same irrational part, #math.equation(block: false, alt: "the square root of 7")[$sqrt(7)$], we can subtract without using a calculator. The rational part of the first number is 3. The rational part of the second number is 8. Using the formula yields #math.equation(block: false, alt: "3 the square root of 7 minus 8 the square root of 7 equals open parenthesis 3 minus 8 close parenthesis times the square root of 7 equals minus 5 the square root of 7")[$3 sqrt(7) − 8 sqrt(7) = ( 3 − 8 ) × sqrt(7) = − 5 sqrt(7)$]. ] ] #examplebox("Example 7")[Adding Irrational Numbers with Similar Irrational Parts][ If possible, add the following irrational numbers without using a calculator. If this is not possible, state why. #math.equation(block: true, alt: "35 π plus 17 π")[$35 π + 17 π$] #solutionbox[ Since these two irrational numbers have the same irrational part, #math.equation(block: false, alt: "π")[$π$], the addition can be performed without using a calculator. The rational part of the first number is 35. The rational part of the second number is 17. Using the formula yields #math.equation(block: false, alt: "35 π plus 17 π equals open parenthesis 35 plus 17 close parenthesis times π equals 52 π")[$35 π + 17 π = ( 35 + 17 ) × π = 52 π$]. ] ] #examplebox("Example 8")[Subtracting Irrational Numbers with Different Irrational Parts][ If possible, subtract the following irrational numbers without using a calculator. If this is not possible, state why. #math.equation(block: true, alt: "19 the square root of 3 minus 5.6 the square root of 7")[$19 sqrt(3) − 5.6 sqrt(7)$] #solutionbox[ The two numbers being subtracted do not have the same irrational part, so the operation cannot be performed. ] ] === Multiplying and Dividing Irrational Numbers Just like any other number that we’ve worked with, irrational numbers can be multiplied or divided. When working with a calculator, enter the operation and a decimal representation will be given. Sometimes, though, you may want to retain the form of the irrational number as a rational part times an irrational part. The process is similar to adding and subtracting irrational numbers when they are in this form. We do not need the irrational parts to match. Even though they need not match, they do need to be similar, such as both irrational parts are square roots, or both irrational parts are multiples of pi. Also, if the irrational parts are square roots, we may need to reduce the resulting square root to lowest terms. When multiplying two square roots, use the following formula. It is the same formula presented during the discussion of simplifying square roots. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ For any two positive numbers #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], #math.equation(block: false, alt: "the square root of a times the square root of b equals the square root of a times b")[$sqrt(a) × sqrt(b) = sqrt(a × b)$]. ] When dividing two square roots, use the following formula. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ For any two positive numbers #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], with #math.equation(block: false, alt: "b")[$b$] not equal to 0, #math.equation(block: false, alt: "the square root of a divided by the square root of b equals the fraction the square root of a over the square root of b equals the square root of the fraction a over b")[$sqrt(a) ÷ sqrt(b) = frac(sqrt(a), sqrt(b)) = sqrt(frac(a, b))$]. ] To multiply or divide irrational numbers with similar irrational parts, do the following: #strong[Step 1:] Multiply or divide the rational parts. #strong[Step 2:] If necessary, reduce the result of Step 1 to lowest terms. This becomes the rational part of the answer. #strong[Step 3:] Multiply or divide the irrational parts. #strong[Step 4:] If necessary, reduce the result from Step 3 to lowest terms. This becomes the irrational part of the answer. #strong[Step 5:] The result is the product of the rational and irrational parts. #examplebox("Example 9")[Dividing Irrational Numbers with Similar Irrational Parts][ Perform the following operations without a calculator. Simplify if possible. + #math.equation(block: false, alt: "3 the square root of 15 divided by open parenthesis 8 the square root of 3 close parenthesis")[$3 sqrt(15) ÷ ( 8 sqrt(3) )$] + #math.equation(block: false, alt: "14.7 the square root of 135 divided by open parenthesis 3 the square root of 5 close parenthesis")[$14.7 sqrt(135) ÷ ( 3 sqrt(5) )$]. #solutionbox[ + In this division problem, #math.equation(block: false, alt: "3 the square root of 15 divided by open parenthesis 8 the square root of 3 close parenthesis")[$3 sqrt(15) ÷ ( 8 sqrt(3) )$], notice that the irrational parts of these numbers are similar. They are both square roots, so follow the steps given above. #strong[Step 1:] Divide the rational parts. #math.equation(block: false, alt: "3 divided by 8 equals the fraction 3 over 8")[$3 ÷ 8 = frac(3, 8)$] #strong[Step 2:] If necessary, reduce the result of Step 1 to lowest terms. The 3 and 8 have no common factors, so #math.equation(block: false, alt: "the fraction 3 over 8")[$frac(3, 8)$] is already in lowest terms. #strong[Step 3:] Divide the irrational parts. #math.equation(block: false, alt: "the square root of 15 divided by the square root of 3 equals the fraction the square root of 15 over the square root of 3 equals the square root of the fraction 15 over 3")[$sqrt(15) ÷ sqrt(3) = frac(sqrt(15), sqrt(3)) = sqrt(frac(15, 3))$] #strong[Step 4:] If necessary, reduce the result from Step 3 to lowest terms. The radicand can be reduced, which yields #math.equation(block: false, alt: "the square root of 5")[$sqrt(5)$]. #strong[Step 5:] The result is the product of the rational and irrational parts, which is #math.equation(block: false, alt: "the fraction 3 over 8 the square root of 5")[$frac(3, 8) sqrt(5)$]. + In this division problem, #math.equation(block: false, alt: "14.7 the square root of 135 divided by open parenthesis 3 the square root of 5 close parenthesis")[$14.7 sqrt(135) ÷ ( 3 sqrt(5) )$], notice that the irrational parts of these numbers are similar. They are both square roots, so follow the steps given above. #strong[Step 1:] Divide the rational parts. #math.equation(block: false, alt: "14.7 divided by 3 equals 4.9")[$14.7 ÷ 3 = 4.9$] #strong[Step 2:] If necessary, reduce the result of Step 1 to lowest terms. This rational number is expressed as a decimal so will not be reduced. #strong[Step 3:] Divide the irrational parts. #math.equation(block: false, alt: "the square root of 135 divided by the square root of 5 equals the fraction the square root of 135 over the square root of 5 equals the square root of the fraction 135 over 5")[$sqrt(135) ÷ sqrt(5) = frac(sqrt(135), sqrt(5)) = sqrt(frac(135, 5))$] #strong[Step 4:] If necessary, reduce the result from Step 3 to lowest terms. The radicand can be reduced, which yields #math.equation(block: false, alt: "the square root of the fraction 135 over 5 equals the square root of 27 equals the square root of 9 times 3 equals 3 the square root of 3")[$sqrt(frac(135, 5)) = sqrt(27) = sqrt(9 × 3) = 3 sqrt(3)$]. #strong[Step 5:] The result is the product of the rational and irrational parts, which is #math.equation(block: false, alt: "4.9 times 3 the square root of 3 equals 14.7 the square root of 3")[$4.9 × 3 sqrt(3) = 14.7 sqrt(3)$]. ] ] #examplebox("Example 10")[Multiplying Irrational Numbers with Similar Irrational Parts][ Perform the following operations without a calculator. Simplify if possible. + #math.equation(block: false, alt: "open parenthesis 19 the square root of 3 close parenthesis times open parenthesis 5.6 the square root of 12 close parenthesis")[$( 19 sqrt(3) ) × ( 5.6 sqrt(12) )$] + #math.equation(block: false, alt: "13 π times 8 π")[$13 π × 8 π$] #solutionbox[ + In this multiplication problem, #math.equation(block: false, alt: "open parenthesis 19 the square root of 3 close parenthesis times open parenthesis 5.6 the square root of 12 close parenthesis")[$( 19 sqrt(3) ) × ( 5.6 sqrt(12) )$], notice that the irrational parts of these numbers are similar. They are both square roots. Follow the process above. #strong[Step 1:] Multiply the rational parts. #math.equation(block: false, alt: "19 times 5.6 equals 106.4")[$19 × 5.6 = 106.4$] #strong[Step 2:] If necessary, reduce the result of Step 1 to lowest terms. This rational number is expressed as a decimal and will not be reduced. #strong[Step 3:] Multiply the irrational parts. #math.equation(block: false, alt: "the square root of 3 times the square root of 12 equals the square root of 3 times 12 equals the square root of 36")[$sqrt(3) × sqrt(12) = sqrt(3 × 12) = sqrt(36)$] #strong[Step 4:] If necessary, reduce the result from Step 3 to lowest terms. The radicand is 36, which is the square of 6. The irrational part reduces to #math.equation(block: false, alt: "the square root of 36 equals 6")[$sqrt(36) = 6$]. #strong[Step 5:] The result is the product of the rational and irrational parts, which is #math.equation(block: false, alt: "106.4 times 6 equals 638.4")[$106.4 × 6 = 638.4$]. Notice that sometimes multiplying or dividing irrational numbers can result in a rational number. + In this multiplication problem, #math.equation(block: false, alt: "13 π times 8 π")[$13 π × 8 π$], notice that the irrational parts of these numbers are the same, #math.equation(block: false, alt: "π")[$π$]. Follow the process above. #strong[Step 1:] Multiply the rational parts. #math.equation(block: false, alt: "13 times 8 equals 104")[$13 × 8 = 104$] #strong[Step 2:] If necessary, reduce the result of Step 1 to lowest terms. That result is an integer. #strong[Step 3:] Multiply the irrational parts. #math.equation(block: false, alt: "π times π equals π squared")[$π × π = π^(2)$] #strong[Step 4:] If necessary, reduce the result from Step 3 to lowest terms. This cannot be reduced. #strong[Step 5:] The result is the product of the rational and irrational parts, which is #math.equation(block: false, alt: "104 π squared")[$104 π^(2)$]. ] ] === Rationalizing Fractions with Irrational Denominators Fractions often represent that some amount is being equally divided into some number of parts. But to conceptualize a fraction in that manner, the denominator needs to be an integer. An irrational number in the denominator interferes with that interpretation of a fraction. Fractions that have denominators that are just the square root of an integer can be altered into fractions with integer denominators using a process called #strong[rationalizing the denominator]. The process relies on the following property of square roots: #math.equation(block: false, alt: "the square root of a times the square root of a equals a")[$sqrt(a) × sqrt(a) = a$] and the following property of fractions: #math.equation(block: false, alt: "the fraction a over b equals the fraction a c over b c")[$frac(a, b) = frac(a c, b c)$] for any non-zero number #math.equation(block: false, alt: "c")[$c$]. Using these two properties, when a fraction has a square root in the denominator, we can eliminate that square root. Multiply the numerator and denominator by that square root from the denominator, #math.equation(block: false, alt: "the fraction a over the square root of b equals the fraction a the square root of b over the square root of b times the square root of b")[$frac(a, sqrt(b)) = frac(a sqrt(b), sqrt(b) × sqrt(b))$]. Then apply #math.equation(block: false, alt: "the square root of a times the square root of a equals a")[$sqrt(a) × sqrt(a) = a$] to the denominator, yielding #math.equation(block: false, alt: "the fraction a the square root of b over the square root of b times the square root of b equals the fraction a the square root of b over b")[$frac(a sqrt(b), sqrt(b) × sqrt(b)) = frac(a sqrt(b), b)$]. Notice that there is no longer a square root in the denominator, which allows for interpreting the fraction as dividing a whole into equal parts. #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Rationalizing_Denominator")[Rationalizing the Denominator] ] #examplebox("Example 11")[Rationalizing the Denominator][ Rationalize the denominator of the following: + #math.equation(block: false, alt: "the fraction 5 over the square root of 7")[$frac(5, sqrt(7))$] + #math.equation(block: false, alt: "the fraction 3 the square root of 6 over 2 the square root of 10")[$frac(3 sqrt(6), 2 sqrt(10))$] #solutionbox[ + The square root in the denominator is #math.equation(block: false, alt: "the square root of 7")[$sqrt(7)$]. In order to rationalize the denominator of #math.equation(block: false, alt: "the fraction 5 over the square root of 7")[$frac(5, sqrt(7))$], we need to multiply the numerator and denominator by #math.equation(block: false, alt: "the square root of 7")[$sqrt(7)$] and simplify. #linebreak() #math.equation(block: false, alt: "the fraction 5 over the square root of 7 equals the fraction 5 the square root of 7 over the square root of 7 times the square root of 7 equals the fraction 5 the square root of 7 over 7")[$frac(5, sqrt(7)) = frac(5 sqrt(7), sqrt(7) × sqrt(7)) = frac(5 sqrt(7), 7)$] #linebreak() The square root is in simplified form, so the final answer is #math.equation(block: false, alt: "the fraction 5 the square root of 7 over 7")[$frac(5 sqrt(7), 7)$]. + The square root in the denominator is #math.equation(block: false, alt: "the square root of 10")[$sqrt(10)$]. #linebreak() #strong[Step 1:] In order to rationalize the denominator of #math.equation(block: false, alt: "the fraction 3 the square root of 6 over 2 the square root of 10")[$frac(3 sqrt(6), 2 sqrt(10))$], we need to multiply the numerator and denominator by #math.equation(block: false, alt: "the square root of 10")[$sqrt(10)$] and simplify. #linebreak() #math.equation(block: false, alt: "the fraction 3 the square root of 6 over 2 the square root of 10 equals the fraction 3 the square root of 6 times the square root of 10 over 2 the square root of 10 times the square root of 10 equals the fraction 3 the square root of 60 over 2 times 10 equals the fraction 3 the square root of 60 over 20")[$frac(3 sqrt(6), 2 sqrt(10)) = frac(3 sqrt(6) × sqrt(10), 2 sqrt(10) × sqrt(10)) = frac(3 sqrt(60), 2 × 10) = frac(3 sqrt(60), 20)$] #strong[Step 2:] The 60 under the square root can be factored into the following factor pairs: #math.equation(block: false, alt: "1 times 60 2 times 30 3 times 20 4 times 15 5 times 12 6 times 10")[$1 × 60 #h(1em) 2 × 30 #h(1em) 3 × 20 #h(1em) 4 × 15 #h(1em) 5 × 12 #h(1em) 6 × 10$] #strong[Step 3:] The largest square factor of 60 is 4, so we simplify the #math.equation(block: false, alt: "the square root of 60")[$sqrt(60)$] in the numerator into #math.equation(block: false, alt: "2 the square root of 15")[$2 sqrt(15)$]. We also cancel any common factors. #math.equation(block: false, alt: "the fraction 3 the square root of 60 over 20 equals the fraction 3 times 2 the square root of 15 over 20 equals the fraction 6 the square root of 15 over 20 equals the fraction 3 the square root of 15 over 10")[$frac(3 sqrt(60), 20) = frac(3 × 2 sqrt(15), 20) = frac(6 sqrt(15), 20) = frac(3 sqrt(15), 10)$] This is completely simplified. ] ] There are occasions when the denominator is irrational but is the sum of two numbers where one or both involve square roots. For instance, #math.equation(block: false, alt: "the fraction 5 over 4 plus the square root of 3")[$frac(5, 4 + sqrt(3))$]. The process used earlier required that the denominator was the square root of a number and would not work here. However, this type of denominator can be rationalized. In order to rationalize such a denominator, we will multiply the numerator and denominator of the fraction by the #strong[conjugate] of the denominator. The conjugate of #math.equation(block: false, alt: "a plus b")[$a + b$] is #math.equation(block: false, alt: "a minus b")[$a #h(0.28em) − b$]. We say that #math.equation(block: false, alt: "a plus b")[$a + b$] and #math.equation(block: false, alt: "a ‒ b")[$a #h(0.28em) ‒ b$] are #strong[conjugate numbers]. So, the conjugate of #math.equation(block: false, alt: "minus 3 plus the square root of 10")[$− 3 + sqrt(10)$] is just #math.equation(block: false, alt: "minus 3 minus the square root of 10")[$− 3 − sqrt(10)$]. But why is this of interest? The reason is because it leads to the #strong[difference of squares] formula, which is used to factor the difference of two squares. Or, for our purposes, in reverse it allows us to eliminate a square root. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ For any two numbers, #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], #math.equation(block: false, alt: "a squared minus b squared equals open parenthesis a minus b close parenthesis open parenthesis a plus b close parenthesis")[$a^(2) − b^(2) = ( a − b ) ( a + b )$]. ] Looking at that formula, you should see that the two factors on the right-hand side of the equals sign are conjugates of one another. So, for our purposes, we’re interested in #math.equation(block: false, alt: "open parenthesis a minus b close parenthesis open parenthesis a plus b close parenthesis equals a squared minus b squared")[$( a − b ) ( a + b ) = a^(2) − b^(2)$]. This tells us that when we multiply #math.equation(block: false, alt: "a plus b")[$a + b$] by its conjugate, we get #math.equation(block: false, alt: "a")[$a$] squared minus #math.equation(block: false, alt: "b")[$b$] squared, or #math.equation(block: false, alt: "a squared minus b squared")[$a^(2) − b^(2)$]. But how is this useful? Let’s return to the fraction above, #math.equation(block: false, alt: "the fraction 5 over 4 plus the square root of 3")[$frac(5, 4 + sqrt(3))$]. The denominator is #math.equation(block: false, alt: "4 plus the square root of 3")[$4 + sqrt(3)$]. Its conjugate is #math.equation(block: false, alt: "4 minus the square root of 3")[$4 − sqrt(3)$]. According to the formula, and letting #math.equation(block: false, alt: "a equals 4")[$a = 4$] and #math.equation(block: false, alt: "b equals the square root of 3")[$b = sqrt(3)$], we see that #math.equation(block: false, alt: "open parenthesis 4 plus the square root of 3 close parenthesis open parenthesis 4 minus the square root of 3 close parenthesis equals 4 squared minus open parenthesis the square root of 3 close parenthesis squared")[$( 4 + sqrt(3) ) ( 4 − sqrt(3) ) = 4^(2) − attach(( sqrt(3) ), t: 2)$]. But #math.equation(block: false, alt: "open parenthesis the square root of 3 close parenthesis squared")[$attach(( sqrt(3) ), t: 2)$] is just 3. That means the product is #math.equation(block: false, alt: "16 minus 3")[$16 − 3$] or 13. This no longer has a square root. We use this to rationalize the denominator. We will also need the #strong[distributive property] of numbers. #notebox("Formula", rgb("#059669"), rgb("#007942"), rgb("#EAF3EC"))[ For any three numbers #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$], #math.equation(block: false, alt: "a times open parenthesis b plus or minus c close parenthesis equals a times b plus or minus a times c")[$a × ( b ± c ) = a × b ± a × c$]. This is called the distributive property. ] #examplebox("Example 12")[Rationalizing the Denominator Using Conjugates][ Rationalize the denominator of #math.equation(block: false, alt: "the fraction 4 over 6 plus the square root of 10")[$frac(4, 6 + sqrt(10))$]. #solutionbox[ #strong[Step 1:] We recognize that the denominator is the sum of two numbers where one or both involve square roots. This means the conjugate can be used to remove the square root from the denominator. #strong[Step 2:] To do so, we multiply the numerator and the denominator each by the conjugate of the denominator. Since the denominator is #math.equation(block: false, alt: "6 plus the square root of 10")[$6 + sqrt(10)$], the conjugate we will use is #math.equation(block: false, alt: "6 minus the square root of 10")[$6 − sqrt(10)$]. #strong[Step 3:] The conjugate is multiplied by the numerator and the denominator. #math.equation(block: true, alt: "the fraction 4 over 6 plus the square root of 10 times the fraction 6 minus the square root of 10 over 6 minus the square root of 10")[$frac(4, 6 + sqrt(10)) × frac(6 − sqrt(10), 6 − sqrt(10))$] #strong[Step 4:] Remembering how a number times its conjugate works, this becomes #math.equation(block: true, alt: "the fraction 4 over 6 plus the square root of 10 times the fraction 6 minus the square root of 10 over 6 minus the square root of 10 equals the fraction 4 times open parenthesis 6 minus the square root of 10 close parenthesis over 6 squared minus open parenthesis the square root of 10 close parenthesis squared")[$frac(4, 6 + sqrt(10)) × frac(6 − sqrt(10), 6 − sqrt(10)) = frac(4 × ( 6 − sqrt(10) ), 6^(2) − attach(( sqrt(10) ), t: 2))$]. #strong[Step 5:] In the numerator, we apply the distributive property. Using it yields #linebreak() #math.equation(block: false, alt: "the fraction 4 times open parenthesis 6 minus the square root of 10 close parenthesis over 6 squared minus open parenthesis the square root of 10 close parenthesis squared equals the fraction 24 minus 4 the square root of 10 over 36 minus 10 equals the fraction 24 minus 4 the square root of 10 over 26")[$frac(4 × ( 6 − sqrt(10) ), 6^(2) − attach(( sqrt(10) ), t: 2)) = frac(24 − 4 sqrt(10), 36 − 10) = frac(24 − 4 sqrt(10), 26)$]. #strong[Step 6:] Notice that the denominator no longer contains a square root. It has been rationalized. If desired, this can then be written as a rational number minus an irrational number, by recalling that #math.equation(block: false, alt: "the fraction a minus b over c equals the fraction a over c minus the fraction b over c")[$frac(a − b, c) = frac(a, c) − frac(b, c)$]. Applying that to the answer, we have #math.equation(block: false, alt: "the fraction 4 over 6 plus the square root of 10 equals the fraction 24 minus 4 the square root of 10 over 26 equals the fraction 24 over 26 minus the fraction 4 the square root of 10 over 26")[$frac(4, 6 + sqrt(10)) = frac(24 − 4 sqrt(10), 26) = frac(24, 26) − frac(4 sqrt(10), 26)$]. #strong[Step 7:] With a bit of cancellation, this reduces to #math.equation(block: false, alt: "the fraction 4 over 6 plus the square root of 10 equals the fraction 24 over 26 minus the fraction 4 the square root of 10 over 26 equals the fraction 12 over 13 minus the fraction 2 the square root of 10 over 13")[$frac(4, 6 + sqrt(10)) = frac(24, 26) − frac(4 sqrt(10), 26) = frac(12, 13) − frac(2 sqrt(10), 13)$]. ] ] #notebox("Video", rgb("#DC2626"), rgb("#DC2626"), rgb("#f7f8fa"))[ #link("https://openstax.org/r/Rationalizing_Denominator")[Rationalizing the Denominator] ] === Key Terms - conjugate numbers - difference of squares - irrational numbers - lowest terms - rationalize the denominator === Key Concepts - Irrational numbers are numbers that cannot be written as an integer divided by another integer. One example is pi, denoted #math.equation(block: false, alt: "π")[$π$]. Another collection of irrational numbers are natural numbers that are not perfect squares. - Some irrational numbers can be written as a rational part multiplied by an irrational part. If two irrational numbers have the same irrational parts, they can be added or subtracted. - When irrational numbers are similar, on can multiply and divide the numbers without a calculator. - Since #math.equation(block: false, alt: "the square root of a times b equals the square root of a times the square root of b")[$sqrt(a × b) = sqrt(a) × sqrt(b)$] , and #math.equation(block: false, alt: "the square root of a divided by the square root of b equals the fraction the square root of a over the square root of b equals the square root of the fraction a over b")[$sqrt(a) ÷ sqrt(b) = frac(sqrt(a), sqrt(b)) = sqrt(frac(a, b))$], products and quotients of square roots can be determined. - Because #math.equation(block: false, alt: "the square root of a squared equals a")[$sqrt(a^(2)) = a$] and #math.equation(block: false, alt: "the square root of a times b equals the square root of a times the square root of b")[$sqrt(a × b) = sqrt(a) × sqrt(b)$], it is possible to simplify square root expressions so the radicand contains no perfect square factors. - When a fraction has an irrational number as its denominator, it is possible to convert the denominator into a rational number using its conjugate. Doing so involves multiplying the numerator and denominator by the conjugate of the denominator, and then applying the difference of squares formula. - With a single square root term - Using conjugate numbers for two term denominators === Formulas - #math.equation(block: true, alt: "the square root of a times b equals the square root of a times the square root of b")[$sqrt(a × b) = sqrt(a) × sqrt(b)$] - #math.equation(block: true, alt: "a times x plus or minus b times x equals open parenthesis a plus or minus b close parenthesis times x")[$a × x ± b × x = ( a ± b ) × x$] - #math.equation(block: true, alt: "the square root of a divided by the square root of b equals the fraction the square root of a over the square root of b equals the square root of the fraction a over b")[$sqrt(a) ÷ sqrt(b) = frac(sqrt(a), sqrt(b)) = sqrt(frac(a, b))$] - #math.equation(block: true, alt: "a squared minus b squared equals open parenthesis a minus b close parenthesis open parenthesis a plus b close parenthesis")[$a^(2) − b^(2) = ( a − b ) ( a + b )$] === Videos - #link("https://openstax.org/r/Philosophy_of_Pythagoreans")[The Philosophy of the Pythagoreans] - #link("https://openstax.org/r/square_root_of_a_number")[Using Desmos to Find the Square Root of a Number] - #link("https://openstax.org/r/Simplifying_Square_Roots")[Simplifying Square Roots] - #link("https://openstax.org/r/Rationalizing_Denominator")[Rationalizing the Denominator]