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Math 3345-Real Analysis — Lecture 02 9/2/05 1. Are There Really

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Math 3345-Real Analysis — Lecture 02 9/2/05 1. Are There Really Math 3345-Real Analysis — Lecture 02 9/2/05 1. Are there really irrational numbers? We got an indication last time that repeating decimals must be rational. I’ll look at another example, and then we’ll expand that into a quick proof. Consider the number x = 845.7256. The first four digits don’t repeat, but after that the sequence of three digits 256 repeats. If we multiply x by 103 = 1000, we get (1) 1000x = 845,725.6256 That’s part of the trick. Now we subtract (2) 1000x − x = 999x = 845,725.6256 − 845.7256 = 844,879.9000 Therefore, we can write x as 844,879.9 8,448,799 (3) x = = , 999 9990 which is a fraction of integers. The repeating decimal x must be a rational number. Theorem 1. A repeating decimal must be rational. (1) Let x be a repeating decimal. (We want to show that x being rational follows from this assumption.) (2) After some point, we must have n digits repeating forever. (This is what it means to be repeating.) (3) y =10nx − x is a terminating decimal. (When you subtract, the repeating decimals cancel out.) y (4) x = 10n−1 is a fraction of a terminating decimal and an integer. (Basic computation.) (5) x must be a fraction of integers. (You can make any terminating decimal into an integer by multi- plying by a power of 10.) (6) x is rational. (Rationals are fractions of integers.) 1.1. So are all integer fractions repeating decimals? As an example, divide 11 into 235 using long division. In your third step, you divide 11 into 40. In your fifth step, you again find yourself dividing 11 into 40. Not surprisingly, you’ll be dividing 11 into 40 in your seventh step. The digits 36 must keep repeating. Theorem 2. Integer fractions are always repeating decimals a Story proof Let’s say we have the integer fraction b . If we do a long division of b into a to convert the fraction to a decimal, we’ll eventually find that our “remainders” are smaller than b. And then we add a zero and divide b into the remainder times 10. The remainder is always smaller than b, so there are only b possible remainders. There must, therefore, be a remainder that occurs more than once. The long division proceeds the same way each time this remainder is encountered, so we will continue to cycle through a fixed sequence of remainders. The result is a repeating decimal (it may be that we repeat only the digit 0). 2. Prime factorizations Every natural number can be factored uniquely into primes. For example, (4) 300 = 2 · 150 = 2 · 2 · 75 = 2 · 2 · 3 · 25 = 2 · 2 · 3 · 5 · 5=223152 =22315270110130 ··· By unique, we mean that the exponent on each particular prime can be only one thing (usually zero). Theorem 3. The prime factorizations of perfect squares have all even exponents (0 is even). All exponents are multiples of 3 for perfect cubes (0 is a multiple of 3), etc. We won’t prove this, but if you look at a few examples, you can see why this is true. 1 2 √ Theorem 4. The number 2 is not rational. √ (1) Suppose 2 is rational. (We’re starting an indirect proof, and hopefully we will arrive at a contra- √diction.) a ∈ 6 (2) 2= b , where a, b Z and b = 0. (Rational numbers can be expressed as fractions.) a2 (3) 2 = b2 . (Basic algebra.) (4) 2b2 = a2. (Basic algebra.) (5) 2b2 and a2 are both integers. (Basic algebra.) (6) 2b2 and a2 have the same unique prime factorization. (Basic fact of life.) (7) For a2 =2α · 3β · 5γ ···, all the exponents must be even. (Since a2 is a perfect square.) (8) For 2b2 =2α · 3β · 5γ ···, α must be odd. (Since b2 is a square and we’re throwing in one more 2.) (9) α must be both odd and even. →← (Repeating the conclusions of the previous two steps.) √ The assumption that 2 is rational√ leads to the contradictory statement that α must be both even and odd. Therefore, it must not be true that 2 is rational. Note that this proof will work for the square root of any prime number. With a little more work, you can show that the square root of any natural number is either going to be a natural number or an irrational number. √ 1. Can 2 be a repeating decimal? Prove it. 3. Homework 02 Before starting problems 1-4, find the prime factorization of 60. 1. What is the exponent on 2? 2. What is the exponent on 3? 3. What is the exponent on 5? 4. What is the exponent on 7? 5. Is 216,000 a perfect cube? For problems 6-8, first find the prime factorization of 216,000. 6. What is the exponent on 2? 7. What is the exponent on 3? 8. For the prime factorization of a perfect cube, the exponents are all multiples of . √ Let’s prove that the cube root of 4 must be irrational. The standard proof idea is to assume that 3 4is rational and show that that leads to contradictory statements. √ Step 1: Suppose that 3 4 is rational. √ 3 a Step 2: Then there are integers a and b such that 4= b . (Definition of rational numbers.) Step 3: b3 · 4=a3. (Basic algebra.) Step 4: 4b3 and a3 must have the same prime factorization. (Fact of life and step 3) Step 5: In this prime factorization, let’s say that α is the exponent on 2. (Notational definition.) Now you finish. 3 9. Since α comes from the prime factorization of a3, it must be a multiple of 10. The exponent on 2 of the prime factorization of b3 must be a multiple of 11. Therefore, α must be a multiple of 3 plus 12. According to problem 11, can α be a multiple of 3? 13. If our original assumption is true, then the conclusions of problems 9 and 12 must also both be true. Are 9 and 12 consistent? √ 14. So is 3 4 rational? My Quiz answer. √ Show 2 is not a repeating decimal. √ (1) Suppose√ 2 is a repeating decimal. (2) Then√ 2 is rational. (By Theorem 1.) (3) But 2 is not√ rational. (By Theorem 4.) (4) Therefore, 2 is both rational and not rational. →←.
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