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3 Spot the mistake All of these proofs have a mistake in them: a clear mathematical error, not just a failure to explain something sufficiently. Briefly explain what the mistake is: a sentence or two should be enough, as long as you clearly point out the error. False theorem: If n is odd, then all multiples of n larger than n are also odd. Bad proof: Suppose for contradiction that n is odd, but all multiples of n larger than n are even. In particular, 3n is even. Therefore the difference 3n-n is an even number minus an odd number, which we know to be odd. But 3n-n=2n, which is clearly even: contradiction! Lying lemma: If n +1 is divisible by 3 and 2n +1 is divisible by 5, then n is even. Wrong proof: Since n + 1 is divisible by 3, there is a k such that n+1 = 3k. Since 2n +1 is divisible by 5, there is a k such that 2n + 1 = 5k. Subtracting the first equation from the second, we get n = 2k, so n is even. Perfidious proposition: All Fibonacci numbers are divisible by 3. Terrible proof: We use strong induction on n to prove that for all n, F, is divisible by 3. n Suppose that F is divisible by 3 for all k

3 Spot the mistake All of these proofs have a mistake in them: a clear mathematical error, not just a failure to explain something sufficiently. Briefly explain what the mistake is: a sentence or two should be enough, as long as you clearly point out the error. False theorem: If n is odd, then all multiples of n larger than n are also odd. Bad proof: Suppose for contradiction that n is odd, but all multiples of n larger than n are even. In particular, 3n is even. Therefore the difference 3n-n is an even number minus an odd number, which we know to be odd. But 3n-n=2n, which is clearly even: contradiction! Lying lemma: If n +1 is divisible by 3 and 2n +1 is divisible by 5, then n is even. Wrong proof: Since n + 1 is divisible by 3, there is a k such that n+1 = 3k. Since 2n +1 is divisible by 5, there is a k such that 2n + 1 = 5k. Subtracting the first equation from the second, we get n = 2k, so n is even. Perfidious proposition: All Fibonacci numbers are divisible by 3. Terrible proof: We use strong induction on n to prove that for all n, F, is divisible by 3. n Suppose that F is divisible by 3 for all k

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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3 Spot the mistake All of these proofs have a mistake in them: a clear mathematical error, not just a failure to explain something sufficiently. Briefly explain what the mistake is: a sentence or two should be enough, as long as you clearly point out the error. False theorem: If n is odd, then all multiples of n larger than n are also odd. Bad proof: Suppose for contradiction that n is odd, but all multiples of n larger than n are even. In particular, 3n is even. Therefore the difference 3n-n is an even number minus an odd number, which we know to be odd. But 3n-n=2n, which is clearly even: contradiction! Lying lemma: If n + 1 is divisible by 3 and 2n + 1 is divisible by 5, then n is even. Wrong proof: Since n + 1 is divisible by 3, there is a k such that n + 1 = 3k. Since 2n + 1 is divisible by 5, there is a k such that 2n + 1 = 5k. Subtracting the first equation from the second, we get n = 2k, so n is even. Perfidious proposition: All Fibonacci numbers are divisible by 3. Terrible proof: We use strong induction on n to prove that for all n, F, is divisible by 3. Suppose that F, is divisible by 3 for all k <n. Then in particular, Fn-1 and Fn-2 are divisible by 3. Let Fn-1= 3a and Fn-2 = 3b. Then Fn = Fn-1+Fn-2= 3a + 3b = 3(a+b), so Fn is divisible by 3.
Transcribed Image Text:3 Spot the mistake All of these proofs have a mistake in them: a clear mathematical error, not just a failure to explain something sufficiently. Briefly explain what the mistake is: a sentence or two should be enough, as long as you clearly point out the error. False theorem: If n is odd, then all multiples of n larger than n are also odd. Bad proof: Suppose for contradiction that n is odd, but all multiples of n larger than n are even. In particular, 3n is even. Therefore the difference 3n-n is an even number minus an odd number, which we know to be odd. But 3n-n=2n, which is clearly even: contradiction! Lying lemma: If n + 1 is divisible by 3 and 2n + 1 is divisible by 5, then n is even. Wrong proof: Since n + 1 is divisible by 3, there is a k such that n + 1 = 3k. Since 2n + 1 is divisible by 5, there is a k such that 2n + 1 = 5k. Subtracting the first equation from the second, we get n = 2k, so n is even. Perfidious proposition: All Fibonacci numbers are divisible by 3. Terrible proof: We use strong induction on n to prove that for all n, F, is divisible by 3. Suppose that F, is divisible by 3 for all k <n. Then in particular, Fn-1 and Fn-2 are divisible by 3. Let Fn-1= 3a and Fn-2 = 3b. Then Fn = Fn-1+Fn-2= 3a + 3b = 3(a+b), so Fn is divisible by 3.
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