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?

  • 2
    $\begingroup$ Welcome to MSE! Not clear what the question is. $\endgroup$
    – VIVID
    Jun 24 '20 at 15:04
  • 3
    $\begingroup$ 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

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.

  • $\begingroup$ 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

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.

  • $\begingroup$ $x \in \mathbb Z$ 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

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 $f(x) = g(x) = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.