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This is a homework question. My class is titled "Formal Mathematical Reasoning and Writing" and we are using Lay's Analysis with an Introduction to Proof. My question comes from section 7: Functions.

Define $f:R \to R$ by $f(x)=x^2$. Find $f^{-1}(T)$ for the following and prove your answer.
$$T=[4, 9)$$

My work thus far:

Claim: $f^{-1}(T)=(-3, -2] \cup [2, 3)$.

Proof: Suppose $a$ is in $f^{-1}(T)$, then by definition of pre-image, $f(a)$ is in $T$. So $a^2$ is in $[4, 9)$ which implies that $4<a^2<9$ Which implies that $2<|a|<3$, thus $a$ is in $(-3, -2] \cup [2, 3)$.

So I believe at this point, I've shown that if $a$ is in the pre-image, then $a$ is in my claimed interval.

I believe I now need to show the converse, that is, if $a$ is in my interval, $a$ is in the pre-image. However, I'm stuck with how to start this part of the proof (and I'm not entirely sure my first half is correct) and where to go... Still fairly green with writing proofs, especially two way proofs.

Any help is appreciated! Thanks!

TMM
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Ben Anderson
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1 Answers1

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Your first half is fine. Suppose $a$ is in your union of intervals. Then $2\le|a|<3,$ so what can you say about $f(a)$?

You can even go "if and only if" all the way. $a\in f^{-1}(T)$ if and only if $f(a)\in T$ if and only if $4\le f(a)<9$ if and only if $4\le a^2<9$ if and only if $2\le|a|<3$ if and only if $a\in(-3,2]\cup[2,3).$ Some of these equivalences follow by definition. Some might require a bit of justification.

Cameron Buie
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  • Thank you, very clear. This got me through the next several examples as well and it became clear towards the end that the iff seems to follow. Slowly but surely, starting to learn this stuff. – Ben Anderson Mar 29 '13 at 15:37