# Thread: Need help on a proof

1. ## Need help on a proof

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.
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!

2. Originally Posted by Selena
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.
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!
Post your attempt for one of them up here. We will help be better able to help you when we see where your troubles lie.

3. Yes of course.

Code:
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)".
I'm pretty sure it's wrong though. :x

4. Originally Posted by Selena
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.
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!
(i) $U(mn)$ is a set of numbers common to $V(m)$ and $W(n)$ where $V(m)$ and $W(n)$ are sets of numbers such that for all positive integers $m$ and $n$, $m= 5j+2$ and $n = 5k+4$ for some positive integers $j$ and $k$

5. Hello Selena!

Originally Posted by Selena
Yes of course.

Code:
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)".
I'm pretty sure it's wrong though. :x
Two hints: $12 = 10 + 2$ and $10 = 5\cdot 2$.

Best wishes,
Sebastian

6. Thanks! The 10+2 part was what it was.
But there's another 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?

7. Hello Selena!

Originally Posted by Selena
But there's another 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?
No, it does not affect the answer. You can use the rule of Universal Elimination and Universal Introduction to prove that the statements which are constituted by switching the n and m are equivalent.

Best wishes,
Seppel

8. Originally Posted by Seppel
Hello Selena!

No, it does not affect the answer. You can use the rule of Universal Elimination and Universal Introduction to prove that the statements which are constituted by switching the n and m are equivalent.

Best wishes,
Seppel
Oh ok I see, thanks!
I'm guessing it only affects if it has one or more "exists" in there right?