Proving One to One Functions: Understanding Strictly Increasing Intervals

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Homework Help Overview

The discussion revolves around proving that a function is one-to-one on an interval if it is strictly increasing on that interval. The original poster seeks assistance in understanding this concept and its proof structure.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using proof by contradiction, questioning how to negate the definition of one-to-one functions. They explore the implications of the strictly increasing property on the relationship between function values and their corresponding inputs.

Discussion Status

Participants have engaged in a productive dialogue, clarifying the steps involved in the proof and confirming the correctness of the reasoning presented. There is a shared understanding of the contradiction approach, though no explicit consensus on a final proof structure has been reached.

Contextual Notes

The original poster expresses uncertainty about the proof process, and hints have been provided regarding the use of contradiction. The discussion remains focused on the theoretical aspects of the problem without delving into specific mathematical methods or solutions.

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Homework Statement



A function from the real numbers to the real numbers is one to one on an interval I if it is strictly increasing on that interval.

Any help please!


Homework Equations




The Attempt at a Solution



I am not quite sure how to prove it, I know that the use of strictly increasing is important as far as if x1 < x2, then f(x1) < f(x2). A hint I was told to use is contradiction.
 
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If you would like to prove it by contradiction, assume there is such a time that f(x) = f(y) where x is not y. Then either x < y or y < x. Now use the strictly increasing property of the function.
 
To prove it by contradiction, negate the definition of one-to-one. Suppose that there are distinct points, a and b, in I, such that f(a) = f(b). You know that either a < b or b < a, right?
 
Great :smile:
 
Okay so:

Assume that f(x1) = f(x2), but x1 does not equal x2, then either x1 < x2 or x2 < x1, since it is strictly increasing, this implies that f(x1) < f(x2) or f(x2) < f(x1), so f(x1) can never equal f(x2), therefore the function must be one to one.

Is this what you are saying?
 
Yes, that is the right idea.

For clarity, instead of saying "f(x1) can never equal f(x2)", just state that "f(x1) < f(x2) or f(x2) < f(x1)" is a contradiction with the fact that f(x1) = f(x2) and thus the function is one-to-one.
 
Alright, thank you SO much for your help!
 
Great, glad I could help.
 

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