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rubikford
02-17-2008, 04:01 AM
I kind of feel dumb about asking this, but here it goes. I have a homework problem that I have to finish and I'm a bit stumped. We're asked to prove that N (set of natural numbers) is numerically equivalent to Z (set of integers). I know that to show that two sets are numerically equivalent, there must be a bijection.

That's where I'm kind stuck. How do I construct the function? I was thinking of a mapping from Z to N, where x goes to lxl (absolute value). Any comments?

Olm
02-17-2008, 04:03 AM
You may want to go to a math forum to ask your question. Here:

PlanetMath (http://planetmath.org/)

Sign up, then post on the appropriate forum.

Dannyb19
02-17-2008, 04:05 AM
It helps if you write it like this:

1, 2, 3, 4, ......

....-3, -2, -1, 0, 1, 2, ..........

Then map 1 to 0, 2 to -1, 3 to 1, 4 to -2, 5 to 2, etc....

You map all the even naturals to the negative integers, and all the odd naturals to the non-negative integers. This is a bijection and shows the cardinality of both sets is equivalent.

rubikford
02-17-2008, 04:09 AM
So, I don't actually have to find a rule or something, like what I did for the problem where I had to show that the set of even integers and the set of integers are numerically equivalent?

zjhtqf
02-17-2008, 04:09 AM
Your way is not right, it's not 1-1.

When N is even, let Z= - N/2

When N is odd, let Z= (N+1)/2

This is a 1-1 mapping.

Dannyb19
02-17-2008, 04:29 AM
Zjhtqf:

A) Mine is 1:1, I do not see your objection
B) Yours does not include zero (unless you include zero in the naturals, which I believe most people do not).

Generally a rule is nice, but there are plenty of proofs and counterexamples where pictures do just fine, but if you want a rule for my description:

When N is even, map to -N/2
When N is odd, map to (N-1)/2

Thus, as I said before:

1 ==> 0
2 ==> -1
3 ==> 1
4 ==> -2
5 ==> 2

And so on....

zjhtqf
02-17-2008, 04:32 AM
To Danny, sorry about the confusion, I referred to the way given by rubikford.

Dannyb19
02-17-2008, 04:33 AM
Oh, ok, no worries zjhtqf, thanks for the clarification!

the_asker
02-17-2008, 04:37 AM
Does this question fall in the "abstract math" category?

pevdoki1
02-17-2008, 04:40 AM
Yeah, I came here expecting some insane abstract algebra proofs :p