Math Help - [SOLVED] Relations

1. [QUOTE=dwsmith;499962][quote=Drexel28;499961]
Originally Posted by dwsmith
Well if it equals zero, the left is odd and the right is odd.

I am confused on how that proves it is an injection though.
We may conclude that $m=m'$ right? And so $f(m,n)=f(m',n')\implies 2^m(2n-1)=2^{m'}(2n-1)=2^m(2n'-1)\implies 2n-1=2n'-1$...soo

2. [quote=Drexel28;499963][quote=dwsmith;499962]
Originally Posted by Drexel28

We may conclude that $m=m'$ right? And so $f(m,n)=f(m',n')\implies 2^m(2n-1)=2^{m'}(2n-1)=2^m(2n'-1)\implies 2n-1=2n'-1$...soo
n=n'

3. [QUOTE=dwsmith;499964][quote=Drexel28;499963]
Originally Posted by dwsmith

n=n'
So!!!

$m=m'\text{ and }n=n'$ and so...

4. [quote=Drexel28;499965][quote=dwsmith;499964]
Originally Posted by Drexel28

So!!!

$m=m'\text{ and }n=n'$ and so...
f is injective if for all a and b in A, if f(a) = f(b), then a = b; that is, f(a) = f(b) implies a = b.

5. [QUOTE=dwsmith;499966][quote=Drexel28;499965]
Originally Posted by dwsmith

f is injective if for all a and b in A, if f(a) = f(b), then a = b; that is, f(a) = f(b) implies a = b.
And $m=m'\text{ and }n=n'\implies (m,n)=(m',n')$...Ta-da!

6. Prove that f is a surjection. This is actually a consequence of the Fundalmental Theorem of Arithmetic, Theorem 8.16.

I don't know what the fact that a number is prime or the product of prime numbers has to be with solving the surjection.

7. Originally Posted by dwsmith
Prove that f is a surjection. This is actually a consequence of the Fundalmental Theorem of Arithmetic, Theorem 8.16.

I don't know what the fact that a number is prime or the product of prime numbers has to be with solving the surjection.
You know that if $n\in\mathbb{N}$ that it may be represented as the product of primes $n=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$, right? Well, except for $2$ every prime is odd. So, we could for the sake of convenience write every integer as $n=2^{\beta}\cdot p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ where $\beta$ may be zero of course. But, as previously mentioned since every other prime is odd and the product of odd numbers is odd it follows that $p_1^{\alpha_1}\cdots p_n^{\alpha_n}\text{ is odd}\implies p_1^{\alpha_1}\cdots p_n^{\alpha_n}=2m+1$ for some $m\in\mathbb{N}$, right? So now what?

8. Originally Posted by Drexel28
You know that if $n\in\mathbb{N}$ that it may be represented as the product of primes $n=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$, right? Well, except for $2$ every prime is odd. So, we could for the sake of convenience write every integer as $n=2^{\beta}\cdot p_1^{\alpha_1}\cdots p_n^{\alpha_n}$ where $\beta$ may be zero of course. But, as previously mentioned since every other prime is odd and the product of odd numbers is odd it follows that $p_1^{\alpha_1}\cdots p_n^{\alpha_n}\text{ is odd}\implies p_1^{\alpha_1}\cdots p_n^{\alpha_n}=2m+1$ for some $m\in\mathbb{N}$, right? So now what?
Well 2m+1 is both the right and left side minus the $2^{\beta}$

9. Originally Posted by dwsmith
Well 2m+1 is both the right and left side minus the $2^{\beta}$
So, every number may be written in the form $\pm2^{\beta}(2m+1)$..so

10. Originally Posted by Drexel28
So, every number may be written in the form $\pm2^{\beta}(2m+1)$..so
Well that product is even unless $\beta=0$

11. Originally Posted by dwsmith
Well that product is even unless $\beta=0$
Does it matter? Isn't that $f(\beta,m)$?

12. Originally Posted by Drexel28
Does it matter? Isn't that $f(\beta,m)$?
Yeah it is f(b,m)

13. Originally Posted by dwsmith
Yeah it is f(b,m)
And a surjection is what?

14. When f(x,y)=z

15. Originally Posted by dwsmith
When f(x,y)=z
And haven't we done that?

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