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Math Help - One to One proof

  1. #1
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    One to One proof

    Let f: R+ --> R+ where R+ is the positive reals, and f(x)=x^4. Prove or disprove that it's one to one.

    Ok, I understand the whole f(a)=f(b) thing down to a=b. But the book says it's not one to one. But wouldn't it be one to one since it's R+ --> R+? Or am I missing something?
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: One to One proof

    Quote Originally Posted by JSB1917 View Post
    Let f: R+ --> R+ where R+ is the positive reals, and f(x)=x^4. Prove or disprove that it's one to one.

    Ok, I understand the whole f(a)=f(b) thing down to a=b. But the book says it's not one to one. But wouldn't it be one to one since it's R+ --> R+? Or am I missing something?
    f(a)=f(b)

    \implies a^4=b^4

    \implies (a^2+b^2)(a+b)(a-b)=0

    \implies a=b or a=-b

    So, f is not one-to-one.
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  3. #3
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    Re: One to One proof

    Quote Originally Posted by JSB1917 View Post
    Let f: R+ --> R+ where R+ is the positive reals, and f(x)=x^4. Prove or disprove that it's one to one.
    Quote Originally Posted by alexmahone View Post
    f(a)=f(b)
    \implies a^4=b^4
    So, f is not one-to-one.
    This function is one-to-one on its domain R^+
    a^4=b^4 implies a=b because a\in R^+~\&~b\in R^+.

    Note that a^4=b^4\text{ if and only if }|a|=|b| but they are both posiive.
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