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Lets say that if f(x) and g(x) are invertible.
1- is (f(x)+g(x)) also invertible? 2- is f(g(x)) invertible too?

for the first one lets say that f(x)=x and g(x)=-x then f(x)+g(x)=x+(-x)=0 and f(x)=0 is not invertible. so my conculusion is that f(x)+g(x) can be uninvertible if f(x) and g(x) are invertible. Am I right?

For the second one I have no clue.

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3 Answers 3

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I have no clue whether the following answers have made headway. As a result, I'll take a stab at this:

For (1), you had to provide an adequate counterexample.

Now, for (2), consider the following: Let $f$ be a function from set $X$ to set $Y$. Likewise, let $g$ be a function from set $Y$ to set $Z$. Thus, $g\circ f$ is a function from set $X$ to set $Z$. We see that $g\circ f$ has an inverse since we can perform $f^{-1}\circ g^{-1}$ to get back to $X$.

Let's decode that. Does it make more sense to think that $f$ is a bridge from place $X$ to place $Y$ and that $g$ is a bridge from place $Y$ to place $Z$? Well, if we have bridges back---that is, $g^{-1}$ being a bridge from place $Z$ to place $Y$ and $f^{-1}$ being a bridge from place $Y$ to place $X$---is it not self evident that to get from place $Z$ to place $X$ we simply take bridge $g^{-1}$ and then bridge $f^{-1}$, and we call this combined route $f^{-1}\circ g^{-1}$? :)


For sake of completeness, here's what I just said in the dry-Algebra language:

Let $f: X\to Y$ and $g: Y \to Z$. Thus, $g\circ f: X \to Z$.

If we have $f^{-1}: Y\to X$ and $g^{-1}: Z \to Y$, we have that $f^{-1}\circ g^{-1}: Z \to X$.

Letting the identity map be $I$, we have $f^{-1}\circ g^{-1}\circ g\circ f=I$. Therefore, $f^{-1}\circ g^{-1}$ is the inverse map of $g \circ f$.

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+1 Very nicely expressed! –  amWhy Nov 20 '12 at 0:16
    
Thank you! I completely fudged up the order of my $f$'s and $g$'s at first, but I have corrected that. –  000 Nov 20 '12 at 0:25
    
You'll get the hang of typesetting; it takes some getting used to! I still can't believe how quickly some users are able to post elaborate responses! –  amWhy Nov 20 '12 at 0:27

You are right about 1. 2 is correct provided $Range(g) \subseteq dom(f)$. In any case it has to be or else $f(g(x))$ would not be well defined.

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HINT: You are right about (1). For (2), suppose that you know that $f\big(g(x)\big)$ is some particular number $y$. Does the invertibility of $f$ guarantee that in principle you can find $g(x)$? And what then does the invertibility of $g$ tell you?

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so is this correct: f(x)=x^3 and g(x)=x^(2/3). Then f(g(x))=(x^(2/3))^3 which is f(g(x))=x^2 which is not invertible. –  user31113 Nov 19 '12 at 21:20
    
@Alex90: Your $g$ isn’t invertible, so this example can’t tell you anything. If $f$ and $g$ are invertible, so is their composition; see if you can follow the hint to see how to explain why. –  Brian M. Scott Nov 19 '12 at 21:27
    
Honestly I dont understand anything from your hint since im a beginner. 1st year in university. Why isnt g invertible? I get only one x for each y. –  user31113 Nov 19 '12 at 21:38
    
@Alex90: $(-8)^{2/3}=(-2)^2=4=2^2=8^{2/3}$, so $g(-8)=g(8)$, and $g$ is not invertible. Look: if $f\big(g(x)\big)=y$, and $f$ is invertible, then isn’t it true that $g(x)=f^{-1}(y)$? Now keep going to find $x$. –  Brian M. Scott Nov 19 '12 at 21:40

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