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I am asked to determine whether a function $f(x) = x^{5} + 1$ is a bijection in $\mathcal{R}$. Proving that the function is one-to-one, I come up with the following:

$\begin{array}{lcl}f(x)& = &f(y)\\x^{5} + 1& = &y^{5} + 1\\x^{5}& = &y^{5}\\x&=&y\end{array}$

However, I am a bit confused about finding whether the function is onto. Based on the definition $\forall x\exists y(f(x) = y)$, I tried the following:

$\begin{array}{lcl}y& = &f(x)\\ y& = &x^{5}+1\\x^5& = &y-1\\x& = &(y-1)^{\frac{1}{5}}\end{array}$

Substituting the right hand side of the equation for $x$ in the above definition for onto functions, I get the following: $$\forall x\exists (y-1)^{\frac{1}{5}}f[(y-1)^{\frac{1}{5}}=y]$$

However, plugging in various values for $y$, the equation always seems to evaluate to the value for $x$. I came to the conclusion that this was not onto, and therefore not a bijection, but the answer in my book says that it is indeed a bijection. What am I doing incorrectly here?

Edit: Silly mistake -- I was simply substituting the value for $x$ as the function body, not accounting for the original function definition.

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2 Answers

The definition of onto is $\forall y\exists x(f(x)=y)$ i.e. every element in the codomain has a preimage in domain.

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You want to prove that $\forall y \exists x(f(x)=y)$. This says what you want. Of course you can interchange the roles of $x$ and $y$, but that will clash with the ordinary notation $y=f(x)$, and may cause the reader, and you, some confusion. You want to show that everybody, like $-47\pi$, is $f$ of something.

And you have in essence done it: For any given $y$, let $x=(y-1)^{1/5}$.

Depending on the type of course you are taking, you may be expected to prove that for any $t$, there is an $s$ such that $s^5=t$, that is, that fifth root always exists. This can be done in various ways. In a calculus course it is done using the Intermediate Value Theorem.

Be careful about the use of quantifiers. In the post you used them in a non-standard way. It is perhaps best at this stage not to use them at all.

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I see my error now -- I forgot to write out the whole function. I was simply making the value for x $(y-1)^{\frac{1}{5}}$ the body of the function. Silly me! –  Dylan Aug 4 '12 at 6:18
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