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the original question

\[\text{Define functions } f \text{ and } g \text{ as follows:}\] \[f(x) = x - \lfloor x \rfloor\] \[g(x) = x - \lceil x \rceil \] \[\text{If } x \in \mathbb{R}\text{, then } \vert f(x) \vert = \vert\ g(x) \vert \]

I hypothesize that the question wants me to prove "for all x in the domain of Real number, absolute of f(x) is equal to absolute of g(x)". Though I am not sure about that so perhaps someone else has another idea?

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    $\begingroup$ Welcome to MSE! Not clear what the question is. $\endgroup$
    – VIVID
    Jun 24 '20 at 15:04
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    $\begingroup$ The two functions are... the same? $\endgroup$ Jun 24 '20 at 15:07
  • $\begingroup$ Calculate $f(1/3)$ and $g(1/3)$. $\endgroup$ Jun 24 '20 at 19:12
  • $\begingroup$ isnt this statement false? $\endgroup$ Jun 24 '20 at 19:30
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You have given two identical functions, therefore, f(x)=g(x) for all x which implies |f(x)|=|g(x)| for all values of x. If, you have copied the question incorrectly, do make sure to upload the correct one.

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  • $\begingroup$ They are not identical. The first is $x$ minus the floor of $x$ and the second is $x$ minus the ceiling of $x$. If $n < x < n+1$ for the integer $n$ then $f(x) = x - n$ and $g(x) =x- (n+1)$. They are not equal. But $|f(x)| = x-n$ and $|g(x)|=(n+1) -x$ and those are not equal so the question is wrong. $\endgroup$
    – fleablood
    Jun 25 '20 at 5:57
  • $\begingroup$ Thats an oops from me :P, excuse my poor eyesight, i read both as the greatest integer function. $\endgroup$
    – Dudeness
    Jul 20 '20 at 4:20
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It looks like f(x) is the (negative) amount rounded up, and g(x) is the amount rounded down, to the next highest and lowest integers respectively.

The equation ${|f(x)| = |g(x)|}$ is not true for all ${x \in \Bbb R}$. If x=1.3, f(x)=-0.7 and g(x)= 0.3.

If the question is to solve (rather than prove) the equation, then ${x = z}$ or ${ z+0.5 : z \in \Bbb Z}$ would do it, I think.

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  • $\begingroup$ $x \in \mathbb Z$ will give you $f(x) =g(x) = 0$. And $x = z+0.5; z\in \mathbb Z$ will give you $f(x) = 0.5$ and $g(x) =-0.5$. A bit more work will tell you that in any other case $|f(x)| = 1-|g(x)|$ $\endgroup$
    – fleablood
    Jun 25 '20 at 5:53
  • $\begingroup$ @fleablood Yes, your answer is much better. Thanks for commenting. $\endgroup$
    – wotnotv
    Jun 25 '20 at 8:28
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If $x$ is an number there is an integer $n$ so that $n \le x < n+1$. That integer is $\lfloor x\rfloor$.

And there is an integer $m$ so that $m-1 < x \le m$. That integer is $\lceil x \rceil$

So $f(x) = x - \lfloor x\rfloor= x - n$ and $0 \le f(x) < 1$.

And $g(x) = x- \lceil x \rceil = x-m$ and $-1 < g(x) \le 0$.

Now $|f(x)| = x- n$ and $|g(x)| = |x-m| = m-x$.

These are not equal. For example of $x = 9.73$ then $f(x) = 9.73 - 9 = 0.73$. And $g(x) = 9.73-10 = -.27$.

Instead what is true is if $x\not \in \mathbb Z$ then $|g(x)| + |f(x)| = 1$.

And if $x \in \mathbb Z$ then $x = \lceil x \rceil = \lfloor x \rfloor$ and $f(x) = g(x) = 0$.

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