3
$\begingroup$

I'm just trying to make sure I have this right:

(b) Give a proof by contradiction of: “If n is an odd integer, then n 2 is odd.”

$n = 2k-1$

$n^2 = (2k-1)^2$

$a = n$ is odd

$b = n^2$ is odd

Since any integer $k$ multiplied by $2$ is even, and we subtract $1$, we get an odd result. That odd result, when squared, is always odd. Therefore, to assume that $n^2$ is even is actually a false statement, giving the following:

$a \rightarrow \neg b$

$TRUE \rightarrow FALSE$

$FALSE$

(c) Give a proof by contraposition of: “If n is an odd integer, then n + 2 is odd.”

$n = 2k-1$

$n+2 = 2k+1$

$a = n$ is odd

$b = n+2$ is odd

Any integer $k$ multiplied by 2 is even, and subtracting 1 is always an odd result. Any odd result + 2 is still odd, so $a$ and $b$ are true. Assuming n+2 is even and n as even would be false statements, so we get the following:

$\neg b \rightarrow \neg a$

$FALSE \rightarrow FALSE$

$TRUE$

$\endgroup$
2
  • 4
    $\begingroup$ Well, they aren'ts proofs by contradiction. They are direct proof that n^2 is odd and then you claim if it were even it'd be false. To do a proof by contradiction START with the assumption that n^2 is even. $\endgroup$
    – fleablood
    Commented Oct 13, 2015 at 2:26
  • $\begingroup$ It is better to write a \iff n is odd which turns into $a \iff n$ is odd , than $a=n $ is odd. $\endgroup$ Commented Oct 13, 2015 at 5:53

2 Answers 2

2
$\begingroup$

Why are you using a contradiction to prove something? $$(2k+1)(2k+1)= \text{ even+even+even}+1!$$ that's the simple way.

$\endgroup$
2
  • $\begingroup$ Sorry to be bringing this valid answer from 9 years ago up, but I think the OP needed to prove the statement by contradiction (not that I disagree with your way of solving it😀). $\endgroup$
    – GSmith
    Commented Jun 10 at 10:51
  • $\begingroup$ Don't mind. I don't understand what is wrong with the way of arguing w.r.t. being odd. Odd=Even+1, thus Odd times Odd equals Even +1. Assume n^2 is even, then n is even, against n is odd. $\endgroup$ Commented Jun 11 at 17:36
2
$\begingroup$

Given: (A) $n$ is odd. Prove (B): $n^2$ is odd.

Your proof is not valid, since in your statement that "that odd result, when squared, is always odd" you have assumed that which was to be proven.

The steps to prove this by contradiction (which is indeed more roundabout than the simplest available proof!) would be:

Assume (B) is false. Then $n^2$ is not odd. By the definition of "odd", there does not exist any integer $m$ such that $n^2 = 2m+1$.

In particular, since for all integer $r$, $2r^2+2r$ ias also an integer, there is no integer $r$ such that $n^2 = 2(2r^2+2r) + 1$.

This implies there is no integer $r$ such that $n^2=4r^2 + 4r + 1$, which implies there is no integer $r$ such that $n=2r+1$ (for if there was, then square both sides to obtain $n^2=4r^2 + 4r + 1$).

So by the definition of "odd", using the symbol $r$ in place of the symbol $m$, $n$ is not odd. But this contradicts (A) -- $n$ is odd -- which was given to be true.

Thus having assumed (B) is false, we have reached a contradiction. Assuming that we accept proofs by contradiction, we have thus proved that (B) is true.

--

You have made the same mistake in the second case, using the theorem that was to be proven, as part of your proof, when you say "any odd result plus two is still odd."

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .