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This was a question on a recent Term Test. I was not sure how to prove the second part of the question and my professor never shows the answers, even after the fact. If someone could please show me how it is solved. I never know how to approach the INT function.

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This is how I answered it, but like I said, I am really not sure: the range of f is (-∞,∞). To find the inverse, we need to consider 2 cases in which x is an integer and x is not an integer: When x is a integer, f(x)=2x so its inverse becomes g(x)=x/2 When x is not an integer the function f becomes f(x)=x+[x]. So there is no inverse for the function when it is not an integer since it cannot form a relation. So f^-1(x)=2x ;x is an integer.

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3 Answers 3

up vote 2 down vote accepted

To prove injectivity and find the inverse all you need is to observe that

$$\{y\}=\{ x+ [x] \}=\{x\} \,.$$ $$[y]=[x+[x]]=[x]+[x]$$ and

$$2[x] \leq y < 2[x]+1 \,.$$

The first two relations show that if $f(x_1)=f(x_2)$ then $x_1,x_2$ have the same integral and fractional parts.

Then, if you want to solve $f(x)=y$ for $y$ in the range of $f$, the last relation shows that $[f(x)]$ is an even integer, so $[y]=2m$. Now you can find the integral and rational part of $x$ from the first two equations.

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This problem needed to be solved without integration because we had not covered the integral yet. –  Dick Jan 25 '13 at 0:29
    
@Dicky "integral part"=greatest integer function. It has nothing to do with integration. –  N. S. Jan 25 '13 at 1:45
    
Sorry about that. –  Dick Jan 25 '13 at 17:37

It might help a lot to try to visualize the function (or get the computer to graph it).

We can show that $x + [x]$ is injective by the fact it is strictly increasing.

By that I mean when $x < y$, $x + [x] < y + [y]$ (because $[x] \le [y]$).

The "range" $R$ is usually called the image: it's simply the unit intervals $[0,1) \cup [2,3) \cup [4,5) \cup \ldots$. To prove this use that $[x]$ is constantly $n$ on $[n,n+1)$.

This proves it has a left inverse $f^{-1} : R \to \mathbb R$.

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range of $f: x+[x]$:

if $x\in[0,1)\;\;\;\;\;\;\;\;x+[x] \in [0,1)$

if $x\in[-1,0)\;\;\;\;\;\;\;\;x+[x] \in [-2,-1)$

Observe that all these intervals can be shifted by 2 to both left and right by adding or subtracting 1 to $x$.

enter image description here

You will see there are gaps that cannot be filled, find a pattern and summarize it.

$f^{-1}$ exists because of injectivity, which comes from monotonicity.

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