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I have been asked to prove the following and am having difficulty:

If $m$ and $n$ are even integers, then so are $m+n$ and $mn$.

My professor has hinted to us to use the definition of an even integer in our proof.

This is my proof for $m+n$ thus far:

  1. Even integers are defined as being divisible by 2.
  2. So, if $m$, is even, using this definition we can rewrite this as $m=2j$.
  3. Similarly, we can rewrite $n$ as $n=2k$.
  4. We now have $m+n=2k+2j$.
  5. Distributivity then allows us to write $2j+2k=2(j+k)$
  6. We now have that $m+n=2(j+k)$.
  7. I now use associativity to create $m+n=(j+k)2$
  8. Next, the definition of divisibility states that 'When $m$ and $n$ are integers, we say $m$ is divisible by $n$ if there exists $j∈ Z$ such that $m=jn$.

  9. This allows me to conclude that, since all even numbers are divisible by 2, that $m+n$ must be an even number.

I am unsure about whether or not my last two steps are correct. Our teacher has hinted to us to use the definition of divisibility, but I am having trouble wrapping my head around a. how to use it, and b. how it is accurate to do so?

Any advice would be much appreciated!

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  • $\begingroup$ That seems good. This depends on your class but your teacher may want to you state that the integers are closed under addition to guarantee that $j + k \in Z$. $\endgroup$ – lordoftheshadows Jan 24 '17 at 4:30
  • $\begingroup$ What you mention in item (8), you used before in (2) and (3), so it's not need to clarify. $\endgroup$ – Nosrati Jan 24 '17 at 4:34
  • $\begingroup$ Nitpick: (9) is a bit imprecise. "Since all even numbers are divisible by 2" together with $m+n$ being divisible by $2$ does not imply that $m+n$ is even. Instead you want to use the implication the other way, i.e. "Since all numbers divisible by 2 are even" and $m+n$ is divisible by $2$, $m+n$ must be even. $\endgroup$ – TMM Jan 24 '17 at 4:41
  • $\begingroup$ Hi, I am having trouble understanding the difference between what I am saying and what you are saying. We haven't gone over logic at all yet so maybe this is where my confusion lies? Thanks :) $\endgroup$ – agra94 Jan 24 '17 at 6:13
  • $\begingroup$ @agra94 What you are essentially saying is "every (even number) has the property that it is (divisible by 2)", and then from the fact that your number is divisible by 2, concluding that it is even. As an analogous example, consider the statement "every (cucumber) has the property that it is (green)", then from the fact that a frog is green you cannot conclude that the frog is a cucumber. For your case you want to use "every (number divisible by 2) has the property that it is (even)", in which case showing that a number is divisible by 2 is indeed enough to show that it is even. $\endgroup$ – TMM Jan 24 '17 at 7:43
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This looks mostly right, well done! As noted in the comments, there is a small error in 9. You wish to use the fact that all numbers that are divisible by $2$ are even to conclude that it's even. Instead, you evoke the fact that all even numbers are divisible by $2$.

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