Claim: all natural numbers are either even or odd. Proof: By induction. Base case:0 is even, since 0 = 2.0. Now suppose that n - 1 is either even or odd, and we must show that n is either even or odd. Case 1: If n - 1 is even, then n - 1 = 2k for some k, so n = 2k + 1, showing n must be odd. Case 2: If n - 1 is odd, thenn - 1 = 2k +1 for some k, so n = 2k + 2 = 2(k+1), showing n must be even. Is this proof correct?
• Yes • No

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QWwhite QWwhite · Answer 1 · 2022-12-22T16:45:57+01:00

No. The proof is wrong.

Here we are asked to prove all Natural numbers are either even or odd.

Natural numbers start from 1 to infinity.

Hence to prove by mathematical induction, we need to take the base case to be n = 1.

here, 1 = 2X0 + 1, therefore 1 is odd

Then we need to assume that for any integer m < n, m is either even or odd.

Now n > 1

Therefore,

n -1 > 0

Hence n - 1 will be a natural number.

Hence by our earlier assumption where any number smaller than is either even or odd

n - 1 has to be either even or odd.

case 1

here n - 1 is odd

if n - 1 is odd then

n-1 + 1 is a sum of two odd numbers, hence n is even

Case 2- If n - 1 is even

then, n-1 + 1 is a sum of an even and n odd number hence n is odd

Therefore n will either be even or odd.

To learn more about Mathematical Induction visit

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