SOLUTION: Prove that the product of two odd numbers is odd, using an indirect proof and a proof by contradiction.

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Question 1070274: Prove that the product of two odd numbers is odd, using an indirect proof and a proof by contradiction.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Here are two odd numbers.
2n+1, and 2n-1, for some integer, n.

We believe that (2n+1)(2n-1) will be 'even', and we can simplify the product expression, through multiplication:


But, is this 'even', or not?

The is undoubtedly EVEN. To that is subtracted the ODD number, 1, which together makes 4n^2-1, the product of the two odd numbers, an ODD number.
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
The "proof" by josgarithmetic" is wrong starting from his second line.

The correct proof is this: 

Let assume that the product of two odd numbers, m and n, is an even number N:  N = m*n.


Then this even number N is a multiple of 2.


The number 2 is a prime number.


Since 2 divides N, it must divide at least one of the factors, n or m.


If 2 divide n, then n is and even number. It contradicts to the original assumption that n is odd.


If 2 divide m, then m is and even number. It contradicts to the original assumption that m is odd.


This/these contradiction/contradictions proves/prove that the product of two odd numbers is an odd number.

QED.   Proved and solved.



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