Can You Help Simplify This Implication Proof?

[Melodia]

Homework Statement

Hi everyone, I need help on proving or disproving this:

Please just show me how to do one of them, and I'd like to try to do the rest on my own. If I don't know then I will post more questions here.

Homework Equations

The Attempt at a Solution

So far, I've interpreted this question this way:
i) for all natural numbers n, {there exists natural number j so that m = 5j + 3 and there exists natural number k so that n = 5k + 4, which works for all natural number m} implies that there exists natural number i so that the product mn = 5i + 2
ii) for all natural numbers m, {there exists natural number i so that m = 5i + 2 and there exists natural number k so that n = 5k + 4, which works for all natural number n} implies that there exists natural number j so that the product mn = 5j + 3
iii) for all natural numbers m, {there exists natural number i so that m = 5i + 2 and there exists natural number j so that n = 5j + 3, which works for all natural number n} implies that there exists natural number k so that the product mn = 5k + 4

But this seems so confusing to me, if anyone could point me in the right direction and show me how to do one of those it would be great!
Thanks in advance!

 

[Dick]

Ok, I think you're interpreting them correctly. Let's try the first one. You've got m=(5j+3) and n=(5k+4). mn=(5j+3)(5k+4). Multiply that out and see if you can write it as 5*(something)+2.

 

[Melodia]

Assume n and m are natural numbers:
	Assume there exists a natural number j and natural number k:
		Assume:
		mn = (5j + 3)(5k + 4):
		   = 25jk + 20j + 15k + 12
		   = 5(5jk + 4j + 3k) + 12
		let i = 5jk + 4j + 3k
		mn = 5i + 12
		Then mn does not equal 5i + 2
	Then m = 5j + 3 and n = 5k + 4
Then V(m) and W(n) together does not imply U(mn)

That is what I have so far for the first one "i)".
It doesn't look right to me though. :x

 

[Dick]

You haven't taken all of the fives out yet. 5i+12=5i+10+2=5i+5*2+2=5(i+2)+2.

 

[Melodia]

Oh! I got it now thanks.
But there's one more problem:
In the "i)", it says "for all n, ..., which works for all m", and the rest are "for all m, ..., which works for all n"; notice that the n and m are switched. Wouldn't that affect the answer?

 

[Dick]

No. "all n and all m" is the same thing as "all m and all n". BTW not all of those are true. For any ones that aren't you just have to find an example of an n and m for which it's not true.

 

[Melodia]

(Lol I just found that you could write symbols with the LaTeX feature)
Oh I see. But if one of the symbols switched to the \exists (there exists one or more), then it would mean different things right?
And here is what I have so far, the "iii)" is disproved:

i)
Assume n and m are natural numbers:
Assume there exists a natural number j and natural number k where
m leaves remainder 3 when divided by j and n leaves remainder 4
when divided by k:
Assume:
mn = (5j + 3)(5k + 4):
= 25jk + 20j + 15k + 12
= 5(5jk + 4j + 3k) + 12
= 5(5jk + 4j + 3k) + 10 + 2
= 5(5jk + 4j + 3k + 2) + 2
let i = 5jk + 4j + 3k +2
mn = 5i + 2
Then mn equal 5i + 2
Then V(m) and W(n) implies U(mn)
Then for all natural m and n, V(m) and W(n) implies U(mn)

ii)
Assume n and m are natural numbers:
Assume there exists a natural number i and natural number k where
m leaves remainder 2 when divided by i and n leaves remainder 4
when divided by k:
Assume:
mn = (5i + 2)(5k + 4):
= 25ik + 20i + 15k + 8
= 5(5ik + 4i + 3k) + 8
= 5(5ik + 4i + 3k) + 5 + 3
= 5(5ik + 4i + 3k + 1) + 3
let j = 5ik + 4i + 3k + 1
mn = 5j + 3
Then mn equal 5j + 3
Then U(m) and W(n) implies V(mn)
Then for all natural m and n, U(m) and W(n) implies V(mn)

Assume n and m are natural numbers:
Assume there exists a natural number i and natural number k where
m leaves remainder 2 when divided by i and n leaves remainder 4
when divided by k:
Assume:
mn = (5i + 2)(5k + 4):
= 25ik + 20i + 15k + 8
= 5(5ik + 4i + 3k) + 8
= 5(5ik + 4i + 3k) + 5 + 3
= 5(5ik + 4i + 3k + 1) + 3
let j = 5ik + 4i + 3k + 1
mn = 5j + 3
Then mn equal 5j + 3
Then U(m) and W(n) implies V(mn)
Then for all natural m and n, U(m) and W(n) implies V(mn)

iii)
Assume n and m are natural numbers:
Assume there exists a natural number i and natural number j where
m leaves remainder 2 when divided by i and n leaves remainder 3
when divided by j:
Assume:
mn = (5i + 2)(5j + 3):
= 25ij + 20i + 15j + 6
= 5(5ij + 4i + 3j) + 6
let k = 5ij + 4i + 3j
mn = 5k + 6
But W(mn) implies that mn = 5k + 4
Then U(m) and V(n) does not imply W(mn)
Then for all natural m and n, U(m) and V(n) does not imply W(mn)

 

[Dick]

Yeah, just use TeX. I'm having a hard time flipping through your "Code" frames. But I think you've got it right. (i) and (ii) are true. (iii) is not. mn=5k+4 and mn=5k+6 can't be true at the same time. Because 6-4 isn't divisible by 5. To put it more simply if you pick m=2 and n=3 then mn=6 doesn't have the form 5k+4.

 

[Melodia]

Oh sorry I changed to quote tags ^^
Thanks for your help =)
If I have more questions later on I'll post in the same thread.
Oh by the way, are there any text editors that has ability to enter these symbols? I can't even find some of these symbols in the character map.

 

[Dick]

Post in a different thread, ok? You'll get a lot more attention that way. I'm probably the wrong one to ask about formatting since I usually butcher the formatting and revert to ascii instead of translating to pretty formatting anyway, if you haven't already noticed.