How Can You Prove a Bijection from [0,1] to [a,b]?

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To prove a bijection from [0,1] to [a,b], an example function like f(x) = c(x - h) can be used, where constants c and h are determined to ensure f(0) = a and f(1) = b. The discussion highlights that functions of the form f(x) = mx + n are indeed bijections when appropriately defined. There is a suggestion to consider different cases based on the values of a and b, but it is emphasized that not all functions from [0,1] to [a,b] are bijections, particularly when considering continuity. The conversation underscores the importance of validating the assumption that all functions are bijections before proceeding with case analysis. Understanding the nature of the functions involved is crucial for solving the problem effectively.
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Homework Statement



I really don't understand the question for this problem, could you please help me out? Thanks so much

1.a,b are some real numbers. Give an example of a bijection from [0,1] to [a,b]
2. Prove that all functions from [0,1] to [a,b] are bijections

Homework Equations


The Attempt at a Solution



I think the example function could be something like Sqrt(1-x) + Sqrt(x). That makes its domain be [0,1]. But I don't know how to make it end up with the value of [a,b]
 
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Try a function like this--f(x) = c(x - h)--with domain restricted to [0, 1]. You'll need to figure out what the constants c and h need to be so that f(0) = a and f(1) = b.
 
Hi, oh yes, your function works. All functions f(x) = mx + n are bijections. Thanks for your help.

About the second one, do you think I should break it into 3 cases. With a,b < 0; a,b > 0; and a <0 ^ b>0?
 
Before looking at your cases, you should first ask yourself whether it's true. "All functions" covers a lot of territory, including functions that are continuous as well as those that are discontinuous.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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