# Functions and inverses proof

1. Consider the subset $\mathbb{R} \times \mathbb{R}$ defined by $D = \{(x,y):|x|+|y| = 1\}$. Describe this set in words. Is it a function?

2. Let $f,g$ be functions, such that $$g\circ f(x) = x,~~\forall x\in~ D(f)\\f\circ g(y) = y,~~\forall y\in D(g)$$ Prove that $g = f^{-1}$.

1. You have a subset $D$ where it is defined as the modulus of $x$ and $y$ when added equals $1$. If I square that, I notice that it will result in $x^2+y^2=1^2$ which then can be associated with the Pythagorean theorem. Is this correct? Can anyone elaborate futher?

2. To have an inverse it has to be one-one. Which means that if $g$ is an inverse of $f$ then $D(g) = R(f) \implies R(g)=D(f)$. So then I have to prove that?

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The Pythagorean stuff is not really right. First note that $|x|+|y|=1$, when $x$ and $y$ are non-negative, just says $x+y=1$. So the first quadrant part of $|x|+|y|=1$ is very familiar. Then reflect in the $x$-axis and/or the $y$-axis. That does not change the absolute values. You should get a "diamond." – André Nicolas Feb 2 '13 at 8:09
@AndréNicolas Kind of confused how exactly is it a diamond? I have a point x and a point y where both added equals 1. So lets say x = 0 and y = 1. How does that make a diamond? – Q.matin Feb 2 '13 at 8:24
You have a lot of points, for example all $(x,y)$ with $x$, $y$ non-negative such that $x+y=1$. That's the line segment joining $(0,1)$ to $(1,0)$. You will also get the line segment joining $(0,1)$ to $(-1,0)$, also the one joining $(-1,0)$ to $(0,-1)$, also $(0,-1)$ to $(1,0)$. Draw it, a diamond (in this case a square turned by $45^\circ$). – André Nicolas Feb 2 '13 at 8:40
@AndréNicolas I see the diamond now thanks. But the first question is not a function correct? – Q.matin Feb 2 '13 at 8:52
That's right. If you have the graph of a function, any vertical line can hit the graph in at most one point. Some vertical lines hit the diamond in two points. Another way: when $x=0$, $y$ could be $1$ or it could be $-1$. But if $y$ is a function of $x$, there is only one $y$-value for any given $x$-value. – André Nicolas Feb 2 '13 at 9:03

For 1. just to verify the basic definition of a function, that is:

If $(x_1,y_1)\in f, (x_1,y_2)\in f$ then $y_1=y_2$.

Now consider $x_1=0$, so you can find $|y|=1$ or $y_1=1,~y_2=-1$ and $y_1\neq y_2$.

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Thanks Babak. Is the modulus sign an absolute value sign or the magnitude? – Q.matin Feb 2 '13 at 8:04
@Q.matin: I assumed it is the first one. Moreover, I agree with Ittay about your secnd approach. Make it clearer. ;) – Babak S. Feb 2 '13 at 8:05
I see from your def of a function Babak that 1 is not a function since $y_1 \ne y_2$ is that correct? – Q.matin Feb 2 '13 at 8:20
@Q.matin: Yes. Furthermore, I want to add something which was covered by Ittay. Try to make a graph of your relation. Make a vertical line to intersect the graph. If it intersects the graph in just one point so the relation is function, otherwise it fails. – Babak S. Feb 2 '13 at 8:26
I like this definition. Usually I see it this spelled out in words (the definition) - but this is mathematically nice and concise! +1 – amWhy Feb 2 '13 at 14:47

To get a feeling for what the shape is formed by $D$ try to find several actual $(x,y)$ points that are in $D$. When you found one try to see if you increase or decrease the $x$ coordinate and still obtain a point in $D$. Can you now figure out what shape $D$ is?

Your formulation of the second question is very vague. It would help if you make it clearer and supply the definition of $f^{-1}$ that you were given.

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$f^{-1}$ is the inverse of $g$ which is stated in the book. Also, for part 1 is it an absolute value sign or magnitude sign? – Q.matin Feb 2 '13 at 8:07
it's an absolute value sign (the absolute value of a number is also called its magnitude). I'm afraid I still have no idea what part 2 is asking. – Ittay Weiss Feb 2 '13 at 8:10
I think I misread your question, when you said the second question is vague do you mean my attempt at a solution is vague or the actual question? – Q.matin Feb 2 '13 at 8:10
the actual question. – Ittay Weiss Feb 2 '13 at 8:11
$D()$ represents the domain. Is that more clear? – Q.matin Feb 2 '13 at 8:13