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Show intervals of real numbers have the same cardinality

  • Thread starter mathcnc
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  • #1
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Homework Statement


Prove the intervals of real numbers (1,3) and (5,15) have the same cardinality by finding an appropriate bijective function of f:(1,3) ->(5,15) and verifying it is 1-1 and onto


Homework Equations


I know there are multiple ways to prove one to one and onto im not sure
which one to use

However, what do i do to prove that cardinality are both infinite?

The Attempt at a Solution


Assume the intervals (1,3) and (5,15)
 

Answers and Replies

  • #2
Hurkyl
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I know there are multiple ways to prove one to one and onto im not sure
which one to use
You're getting ahead of yourself. You can't prove that a function is one to one and onto unless you have a function... and you haven't selected one yet!


However, what do i do to prove that cardinality are both infinite?
I think the easiest way would be to find a subset that is easy to prove infinite. But that wasn't what was asked of you... do you have an idea about how to use such a fact?
 
  • #3
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i don't know im so confused!
 
  • #4
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(1,3) and (5,15) are pretty similar; one is just "longer" than the other. Try imagining each as a line segment without endpoints. What's a simple way to bijectively map points on one line segment to the other? Think about scaling and translation.
 
  • #5
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so liek i would say that
(1,2) is a subset and it have infinite cardinality and so (1,5) does?
im sorry im jsut having major issues with going about how to prove this
 
  • #6
Dick
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so liek i would say that
(1,2) is a subset and it have infinite cardinality and so (1,5) does?
im sorry im jsut having major issues with going about how to prove this
Ok, if you are that confused think about f(x)=5*x. Can you show that's a 1-1 and onto map from (1,3) to (5,15)? That's what you need to show they have the same cardinality.
 
  • #7
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It is true that (1,2) and (1,5) are both infinite in size. Likewise, it is true that (1,3) and (5,15) are infinite in size. But that does not immediately mean that they all have the same size.

Consider your original question again. You want to prove that the intervals (1,3) and (5,15) are the same size i.e. have the same cardinality. Therefore, you need to show that each element in (1,3) has an unique corresponding element in (5,15). Likewise, every element in (5,15) has an unique corresponding element in (1,3).

Another way to think about it, and why your question wants you to "find an appropriate bijective function of f:(1,3) -> (5,15)" is that there is a function that maps every element of (1,3) to an unique element of (5,15). But for the two sets to have the same size, every element of (5,15) must also map to an unique element of (1,3). Why are we guaranteed this if f is one-to-one and onto (i.e. bijective)?
 

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