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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 Molly
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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Answered by
drwls
This kind of "new wave" math teaching will probably make it even harder to learn algebra. I hope not.
I had never heard of a bijection before. So I looked up this website:
http://en.wikipedia.org/wiki/Bijection
It seems obvious that for every x there is a unique f(x), and for every f(x) there is a unique x. That fulfills the requirement for a bijection.
I had never heard of a bijection before. So I looked up this website:
http://en.wikipedia.org/wiki/Bijection
It seems obvious that for every x there is a unique f(x), and for every f(x) there is a unique x. That fulfills the requirement for a bijection.
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