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## #1 2006-04-18 14:28:21

Kazy
Member
Registered: 2006-01-24
Posts: 37

### Numerically Equivalent Proof

I need to prove the following:

a) Prove that every closed interval [a,b] is numerically equivalent to [0,1].
b) Prove that any two closed intervals [a,b] and [c,d] are numerically equivalent.
c) Prove that any two open intervals (a,b) and (c,d) are numerically equivalent.
d) Let [a,b] be a closed interval and (c,d) be an open interval. Prove that [a,b] and (c,d) are numerically equivalent.

I'm guessing all of these are proven pretty much the same way, but I don't even know where to start. Can anyone help?

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## #2 2006-04-18 17:30:15

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: Numerically Equivalent Proof

Can you define Numerically Equivalent?

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #3 2006-04-19 04:27:39

Kazy
Member
Registered: 2006-01-24
Posts: 37

### Re: Numerically Equivalent Proof

Let A and B be sets contained in some universal set U. We say that A and B are numerically equivalent if there exists a bijection f: A->B.

That help?

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## #4 2006-04-19 04:54:42

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: Numerically Equivalent Proof

Yep.

a.
It should be obvious that the easiest map to make has a going to 0 and b going to 1.  So f(a) = 0 and f(b) = 1.

Using this logic, we can then say that mid points map to mid points.  So f((a-b)/2) = 1/2.  Continue this logic, and you will eventually get every point in [a, b].

So doing some guess and checking, I come up with: f(c) = (c - a)/(b - a)

Note that a and b are constants, thus, this is just a linear map.  Your teacher may or may not allow you to use the fact that all linear maps are bijective.  Also note that it fails for when a = b, as expected.

b follows the same exact logic as a.  But, instead of having from 0 to 1, you have from c to d.

I'll take a look at c and d later.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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