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Irrational square roots MichaΜ·l Misiurewicz ([email protected]), Indiana University - Purdue University Indianapolis, Indianapolis, IN β In the classroom. Suppose you show your students the standard proof that 2 is irrational: Proof by contradiction. Suppose that β 2 is rational. Then we can write β 2 = π/π for some coprime positive integers π and π. This means that π2 = 2π2 . Thus, π2 is even, and therefore π is even. Hence, π = 2π£ for some positive integer π£. We get 4π£ 2 = 2π2 , so 2π£ 2 = π2 . By the same argument, this implies that π is even, which β contradicts the assumption that π and π are coprime. Therefore 2 is irrational. Of course, the same proof works for every prime π. To prove that β π is irrational, replace 2 by π and βevenβ by βdivisible by π.β You wonder whether your students understand this principle. After all, it is not clear that for a student the notions of βevenβ and βdivisible by πβ are similar. After β β you prove the irrationality of 2 in class, you ask your students to prove that 3 is irrational. One studentβs proof starts by distinguishing the cases of π and π even or odd. This cannot be correct, you think, but before you write a big red zero on the paper, you read the rest of the proof. Suppose that β 3 is rational. Then we can write β 3 = π/π for some coprime positive integers π and π. This means that π2 = 3π2 . If π is even, then π2 is even, so π is even β a contradiction. If π is even, then 3π2 is even, so π is even β another contradiction. Thus, both π and π are odd, so π = 2π£ + 1 and π = 2π€ + 1 for some nonnegative integers π£, π€. We get (2π£ + 1)2 = 3(2π€ + 1)2 , so 2π£ 2 + 2π£ = 6π€ 2 + 6π€ + 1. The left-hand side is even, while the right-hand side is odd β β a contradiction. Therefore 3 is irrational. 1 2 The proof is correct! Your student has earned a perfect score instead of 0, but did not learn what you wanted to teach. Next time you are teaching this course, you do not repeat your mistake! You ask yourself: for what prime numbers does this βeven-oddβ proof work? You easily see that it works for all integers congruent to 3 modulo 4. It is clear what you have to do. The smallest prime number larger than 2 but not congruent to 3 modulo 4 is 5. β Thus, you ask your students to prove that 5 is irrational. One student submits this solution: Suppose that β 5 is rational. Then we can write β 5 = π/π for some coprime positive integers π and π. This means that π2 = 5π2 . If π is even, then π2 is even, so π is even β a contradiction. If π is even, then 5π2 is even, so π is even β a contradiction. Thus both π and π are odd, so π = 2π£ + 1 and π = 2π€ + 1 for some nonnegative integers π£, π€. We get (2π£ + 1)2 = 5(2π€ + 1)2 , so π£ 2 + π£ = 5π€ 2 + 5π€ + 1. Now, π£ 2 + π£ = π£(π£ + 1). Either π£ or π£ + 1 is even, so π£ 2 + π£ is even. Similarly, π€ 2 + π€ is even. Therefore the left-hand side of the equality π£ 2 + π£ = 5π€ 2 + 5π€ + 1 is even, β while the right-hand side is odd β a contradiction. Therefore 5 is irrational. Not again! This time, as you can again easily check, the proof works for all π congruent to 5 modulo 8. You are desperate. What can you do? To check whether β your students understand the idea of the proof of irrationality of 2, what can you ask them to prove? Problem. For which odd numbers π can we prove that β π is irrational using the βeven-oddβ method? Let us translate the βeven-oddβ method into the rigorous language of mathematics. β Assume that π = π/π with (π, π) = 1. Then π2 = ππ2 . (1) 3 Depending on whether π is even or odd, we write π = 2π£ or π = 2π£ + 1; depending on whether π is even or odd, we write π = 2π€ or π = 2π€ + 1 We substitute those expressions in (1). This gives us 4 possibilities. Some of them lead immediately to a contradiction (an even number equals an odd number), some of them may require further analysis. βFurther analysisβ means that we continue the same procedure with π£, π€ instead of π, π, that is, π£ may be even or odd and π€ may be even or odd. Repeating this procedure β times amounts to considering 22β possibilities (although some may be ruled out earlier in the process): π β‘ π (mod 2β ) for some π β {0, 1, . . . , 2β β 1}, and π β‘ π (mod 2β ) for some π β {0, 1, . . . , 2β β 1}. In fact, many cases are ruled out because π and π are odd. Thus, our proof is successful if, for some β β₯ 1, all possibilities are ruled out, that is, the equation π 2 β‘ ππ 2 (mod 2β ) (2) has no solutions in odd π, π . When solutions exist, we continue to β + 1. Since π is odd, it has an inverse modulo 2β . Multiplying both sides of (2) by the square of this inverse, we get π‘2 β‘ π (mod 2β ). We see that our problem is equivalent to the following: For which odd π is there an β β₯ 1, such that for every π‘ we have π‘2 ββ‘ π (mod 2β )? Solution. As we have noticed already, the βeven-oddβ method works if π β‘ 3, 5, 7 (mod 8). We can also see this by observing that π‘2 β‘ 1 (mod 8) for all odd π‘. We will show that it does not work if π β‘ 1 (mod 8). That is, we show that if π β‘ 1 (mod 8) then for every β β₯ 1 there is π‘ such that π‘2 β‘ π (mod 2β ), and the βeven-oddβ proof will never terminate. Let π β‘ 1 (mod 8). Then for π‘ = 1 we get π β‘ π‘2 (mod 2β ) for β = 1, 2, 3. Now we apply induction. Namely, we show that if β β₯ 3 and π β‘ π‘2 (mod 2β ) then either π β‘ π‘2 (mod 2β+1 ) or π β‘ (π‘ + 2ββ1 )2 (mod 2β+1 ). Indeed, if π β‘ π‘2 (mod 2β ) then either π β‘ π‘2 (mod 2β+1 ) (and we are done), or π β‘ π‘2 + 2β (mod 2β+1 ). In the latter 4 case, since 2β β 2 β₯ β + 1 and π‘ is odd, (π‘ + 2ββ1 )2 β‘ π‘2 + 2β π‘ + 22ββ2 β‘ π‘2 + 2β π‘ β‘ π‘2 + 2β + 2β (π‘ β 1) β‘ π‘2 + 2β β‘ π (mod 2β+1 ). Thus, by induction we prove that for every β β₯ 1 there is π‘ such that π β‘ π‘2 (mod 2β ). This solves our problem. Theorem. The βeven-oddβ method of proving that β π is irrational works for an odd π if and only if π β‘ 3, 5, 7 (mod 8). Conclusion. To check whether your students understand the proof of irrationality of β 2, ask them to prove that β 17 is irrational.