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If two integers have opposite parity, then their product is even.

Proof Method: Direct Proof

If two integers have opposite parity, then one is even and the other is odd.

Suppose: $a$ is an even integer and $b$ is an odd integer, then by definition of even and odd integers

$$a = 2m, \quad b = 2n+1,$$ while $m$ and $n$ are integers.

$$ ab = 2m(2n+1)= 4mn+2m = 2(2mn+m) $$ Let $c = 2mn+m$ be an integer, then $ab=2c$ is even

Therefore, the product of two opposite parity integers is even

Thank You!

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    $\begingroup$ One suggestion for improvement: Prove that $2mn + m$ is an integer. Since $2, m, n$ are integers and the product of integers is an integer, $2mn$ is an integer. Since $2mn$ and $m$ are integers and the sum of integers is an integer, $2mn + m$ is an integer. $\endgroup$ – N. F. Taussig Feb 21 '18 at 1:35
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    $\begingroup$ For anyone visiting from Stack Overflow who is as confused as I was at first, parity in mathematics is not the same as parity in computing. In math, parity is whether an integer is even or odd. In computing, parity is whether the number of 1 bits in the binary representation of the integer is even or odd. For example, in computing, 3 is binary 11 with even parity. 7 is binary 111 with odd parity. And their product, 21, is binary 10101, so it is both odd and has odd parity. $\endgroup$ – Michael Geary Feb 21 '18 at 7:07
  • $\begingroup$ Another, possible proof would be : Duh. $\endgroup$ – Eric Duminil Feb 21 '18 at 9:15
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Yes of course that's correct.

We can also observe that

  • if $a\in \mathbb{N}$ is even $\implies 2\,|\,a\,$ and $\,\forall b \in \mathbb{N} \quad 2\,|\,ab \implies ab$ is even.
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    $\begingroup$ Use \mid for for better spacing of the divisibility symbol $\mid$. For example, instead of $2|ab$, you could generate $2\mid ab$. Notice the better spacing. $\endgroup$ – Mr Pie Feb 21 '18 at 1:20
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    $\begingroup$ Thanks for the suggestion , I fix it! $\endgroup$ – gimusi Feb 21 '18 at 1:29
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    $\begingroup$ No problem :) ${}$ $\endgroup$ – Mr Pie Feb 21 '18 at 3:39
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Euclid's Lemma asserts that if a prime number $p$ divides the product $ab$ of two integers $a$ and $b$, then $p$ must divide at least one of those integers $a$ and $b$.

Since $2$ is a prime number, we can let $p = 2$. We now want to show that for integers $a$ and $b$ that have an opposite parity, $2$ must divide the product $ab$.

When two numbers have an opposite parity, then one of those numbers is even and the remaining number is odd. The definition of an even number is such that the number is divisible by $2$. The contrary is the definition of an odd number. This is why we let $p = 2$ in the first place.

Since either $a$ or $b$ must be even, the remaining integer consequently being odd, then $2$ must divide either $a$ or $b$. Therefore, $2$ must also divide $ab$ which makes that product an even number.

Go here for a proof of Euclid's Lemma.

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    $\begingroup$ -1: you are not using "Euclid's lemma" at all. You are using its converse (if one of a or b is even then ab is even), which is almost trivially true. $\endgroup$ – Tashi Walde Feb 21 '18 at 2:09
  • $\begingroup$ @TashiWalde Yes sorry. I forgot to add why we let $p = 2$. Otherwise, you are correct. $\endgroup$ – Mr Pie Feb 21 '18 at 3:40

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