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I am asked:

Determine if the following statement is true or false by considering its contrapositive. Prove the statement if it is true.

If $x$ and $y$ are two integers for which $3(x-y)$ is even, then $x$ and $y$ have the same parity.

Here's my attempt:

Proof: (Contrapositive) Assume that $x$ and $y$ do not have the same parity, i.e. $x$ is odd and $y$ is even, or $x$ is even and $y$ is odd.

First, suppose that $x$ s off such that $x=2k+1,k\in \mathbb{Z}$ and $y$ is even such that $y=2l,l\in \mathbb{Z}$. Then $3(x-y)=3(2k+1-2l)=6k+3-6l=(6k-6l+1)+1=2(3k-3l+1)+1$. Since $k\in \mathbb{Z}$ and $l\in \mathbb{Z}$, $3k\in \mathbb{Z}$ and $3l\in \mathbb{Z}$. Thus, $2(3k-3l+1)+1$ is odd.

Alternatively, suppose that $x$ is even and $y$ is odd such that $x=2m,m\in \mathbb{Z}$ and $y=2n+1,n\in \mathbb{Z}$. Then $3(x-y)=3(2m-(2n+1))=3(2m-2n-1)=6m-6n-3=(6m-6n-2)-1=2(3m-3n-1)-1$. Since $m\in \mathbb{Z}$ and $n\in \mathbb{Z}$, $3m\in \mathbb{Z}$ and $3n\in \mathbb{Z}$. Thus, $2(3m-3n-1)-1$ is odd.

In either case we have shown that $x \text{and} y \text{do not have the same parity})\implies (3(x-y)\text{is odd})$ holds true, which is the contrapositive of the original statement. Therefore, the original statement must be true. $\blacksquare$

Brandon O'Neil
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1 Answers1

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Your proof is good but complicated.

You want to prove: If $3(x-y)$ is even then $x$ and $y$ have same parity.

So the contraposition (which you never actually stated) is : If $x$ and $y$ have opposite parity, the $3(x-y)$ is odd.

Pf:

Lemma 1: the sum or difference of two number of opposite parity is odd.

So $x -y$ is odd.

Lemma 2: the product of two odd numbers is odd.

Lemma 3: $3$ is an odd number.

So $3$ is odd and $x -y$ is odd so $3(x-y)$ is odd.

Done

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Proof of Lemma 1: Let the numbers be $2k +1$ and $2j$ then the sum is $2(k+j) + 1$ is odd. Difference are just a sum with the a negative of on of the numbers which has the same parity so the result follows. If that seems too glib $\pm[2k+ 1 - 2j] = \pm 2(k-j) \pm 1$ is odd.

Proof of Lemma 2: Let $2k+1; 2j+1$ be two odd numbers. Their product is $4kj + 2k + 2j + 1 = 2(2kj + k + j) + 1$ is odd.

Proof of Lemma 3: $3 = 2*1 + 1$. Okay, that was a tongue in cheek .....

fleablood
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