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Suppose that R is the set of real numbers. Let f: R--> R be defined by f(x)= mx+b, where m and b are real numbers and m is nonz...Asked by Adam
Suppose that R is the set of real numbers. Let f: R--> R be defined by f(x)= mx+b, where m and b are real numbers and m is nonzero. Prove that f is a bijection.
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MathMate
A bijective function is a function that is both one-to-one (injective) and onto (surjective).
To prove that f(x)=mx+b is bijective requires two parts, f(x) is one-to-one and f(x) is onto.
Here are some example references that can help you prove that f(x) is one-to-one (injective), and f(x) is onto (surjective). If you can prove these two properties, then f(x) is bijective.
Definitions of injection (one-to-one) and surjection (onto):
http://en.wikipedia.org/wiki/Injective_function
http://en.wikipedia.org/wiki/Surjective_function
Example proofs:
http://www.math.csusb.edu/notes/proofs/bpf/node4.html
http://www.math.csusb.edu/notes/proofs/bpf/node5.html
You're welcome to post your proof for review.
To prove that f(x)=mx+b is bijective requires two parts, f(x) is one-to-one and f(x) is onto.
Here are some example references that can help you prove that f(x) is one-to-one (injective), and f(x) is onto (surjective). If you can prove these two properties, then f(x) is bijective.
Definitions of injection (one-to-one) and surjection (onto):
http://en.wikipedia.org/wiki/Injective_function
http://en.wikipedia.org/wiki/Surjective_function
Example proofs:
http://www.math.csusb.edu/notes/proofs/bpf/node4.html
http://www.math.csusb.edu/notes/proofs/bpf/node5.html
You're welcome to post your proof for review.
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