Proving that the product of an even integer and any positive integer is even

[sonadoramante]

Proof:
Let a be a even positive integer of the form a=2m & b of the form b=2n (This is where b is a even positive integer)
ab = 2m*2n
= 2(mn)
= Let k = mn
= 2k
Therefore, ab is even.

Let a be a even positive integer a=2m & b be a odd positive integer b = 2n+1
ab = (2m)*(2n+1)
= 4mn + 2m
= 2(mn+m)
= Let k = (mn+m)
= 2k
Hence ab is even.

 

[PeroK]

Proof:
Let a be a even positive integer of the form a=2m & b of the form b=2n (This is where b is a even positive integer)
ab = 2m*2n
= 2(mn)
= Let k = mn
= 2k
Therefore, ab is even.

Let a be a even positive integer a=2m & b be a odd positive integer b = 2n+1
ab = (2m)*(2n+1)
= 4mn + 2m
= 2(mn+m)
= Let k = (mn+m)
= 2k
Hence ab is even.

Okay, but can you improve on that?

PS did the question specify that $a$ is positive?

 

[mfb]

I moved your threads to our homework section.

2m*2n
= 2(mn)

Check this step.

You can simplify the overall proof quite a bit. Does it matter if b is even or odd?

 

[sonadoramante]

I moved your threads to our homework section.
Check this step.

You can simplify the overall proof quite a bit. Does it matter if b is even or odd?

It wasn't specified, however, I was trying to prove it for both cases (if b is even or odd positive integer)

 

[sonadoramante]

Okay, but can you improve on that?

PS did the question specify that a is positive?

No! My bad. It didn't specify in the question that a is positive.

 

[mfb]

Oh sure, you need to prove it for all b, even or odd, but you can do so in a single step. There is no need to distinguish the two cases.

 

[sonadoramante]

Oh sure, you need to prove it for all b, even or odd, but you can do so in a single step. There is no need to distinguish the two cases.

After feedback.

Let p be a even integer & q be any positive integer.
Given that even number*even number is even
& even number*odd number is even, then..

p=2m
pq= 2mq
pq= 2(mq)
Let k=mq
pq=2k
Hence pq is even. :)

 

[PeroK]

After feedback.

Let p be a even integer & q be any positive integer.
Given that even number*even number is even
& even number*odd number is even, then..

p=2m
pq= 2mq
pq= 2(mq)
Let k=mq
pq=2k
Hence pq is even. :)

That's okay, but (in my opinion) you are not emphasisng the key points.

First, as $p$ is even $p = 2m$ for some integer $m$.

Second, $pq = 2(mq)$. Now, whether or not you introduce $k = mq$, the key point is that $mq$ is an integer. Hence $pq$ is even.

 

[mfb]

Given that even number*even number is even
& even number*odd number is even, then..

That's what you want to prove, you can't take it as given.

 

[sonadoramante]

That's okay, but (in my opinion) you are not emphasisng the key points.

First, as $p$ is even $p = 2m$ for some integer $m$.

Second, $pq = 2(mq)$. Now, whether or not you introduce $k = mq$, the key point is that $mq$ is an integer. Hence $pq$ is even.

Thanks a lot.