Showing two sets are equivalent

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    Equivalent Sets
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Homework Help Overview

The discussion revolves around demonstrating the equivalence of the set of all real numbers and the set of all positive real numbers, specifically through the identification of a bijection between the two sets.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore various functions as potential bijections, including the tangent function and the exponential function. There is a focus on the need for a one-to-one mapping to establish equivalence.

Discussion Status

Some participants have suggested functions that may serve as bijections, while others have pointed out flaws in those suggestions. There is an ongoing exploration of different functions, with some guidance being offered regarding the use of exponential functions.

Contextual Notes

Participants are working under the premise that a bijection is necessary to prove the equivalence of the two sets, and there is some confusion regarding the definitions and properties of the functions being considered.

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Homework Statement



Show that R \approx R+ , that is, the set of all real numbers is equivalent to the set of all positive real numbers


Homework Equations



The only relevant equation is finding one such that F:R\rightarrowR+ is a bijection.

The Attempt at a Solution



I've attempted to use the tangent function as a bijection, but that doesn't work. This isn't for homework but for a test review sheet, so all guidance is gladly appreciated.
 
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Equivalent? You could probably show that the sets are isomorphic by finding an isomorphism. The function:
F: \mathbb{R} \to \mathbb{R}^+
F:x \to x^2
would map the real numbers into positive real numbers. I should think F is an isomorphism over these sets.
 
But that isn't a bijection as its not one-to-one. For example, -5 maps to 25 and 5 maps to 25. If I can find a bijection from R to R+ ,then I've proven they are equivalent sets. Just can't find the darn bijection.
 
Hi hwill205:smile:

You should think of exponential functions...
 
Damn, y=ex works. Thanks a lot man. Can't believe I didn't think of that.
 
hwill205 said:
But that isn't a bijection as its not one-to-one. For example, -5 maps to 25 and 5 maps to 25. If I can find a bijection from R to R+ ,then I've proven they are equivalent sets. Just can't find the darn bijection.

Oh, of course. That + sign made me think of positive real numbers, and I messed up codomain and domain. :(
 

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