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  1. Consider the set $A={(x^2, x):x \in R}$. Is this a function from $R$ to $R$?

I know it will be a function if there is a unique output per input, but I've never seen a function formatted like this. Is the $x^2$ the independent value and $x$ the dependent?

  1. A function $f:Z \rightarrow Z x Z$ is defined as $f(n) = (2n, n+3)$. Verify whether this function is injective and whether it is surjective.

Here I'm confused how to prove either injective and surjective. I can't see an $n$ such that I get the same output making it not injective, thus I think it's injective...but not sure how to show it.

I don't think it's surjective because $f(n)$ does not map to $(1, 0)$ for example. But if something were surjective how would I prove it? Thanks for any help!

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    $\begingroup$ A function $f:A\to B$ is "really" a subset $f$ of $A\times B$ with the following property: for each $x\in A$ there is one and only one $y\in B$ such that $(x,y)\in f$. $\endgroup$
    – Pedro
    Commented Oct 3, 2013 at 2:22

2 Answers 2

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For your first question, suppose that we had two elements of $A$ with the same coordinate: that is, we had $(a, b) \in A$ and $(a, c) \in A$. Then by definition of $A$, we have that $b^2 = a = c^2$, so $b^2 = c^2$. So does it follow that $b = c$? (It doesn't.)

For the second, you are correct that it is injective: For if $f(n) = f(m)$, then $$(2n, n + 3) = (2m, m + 3) \implies n + 3 = m + 3 \implies n = m$$ You're also correct that it's not onto, since it never hits $(1, 0)$. If it were onto, then you'd select an arbitrary element of $\Bbb{Z} \times \Bbb{Z}$ and find an element of $\Bbb{Z}$ mapping to it.

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This is a relation between the square real numbers and the real numbers. Each square real number $x^2\to x $, Notice that $9\to 3$ and $9\to -3$. Therefore, this cannot be a function.

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