Question

\[f:\mathbb{R} \rightarrow \mathbb{R} \] is continious with \[Im(f) \subset \mathbb{Z}\] then \[f\] is constant. Give a short reason why or give an example why its not.

Answer 1

@jim_thompson5910 @zepdrix
any ideas ?

Answer 2

if the language is not clear enough tell me, so i can try to make it more preciser, my english is not very well in some cases its possible i use wrong words, for example i am suspecious about constant in german its konstant hope has same meaning in english :)

Answer 3

@Directrix @cwrw238

Answer 4

@oldrin_batuku @satellite73 @amistre64

Answer 5

@SithsAndGiggles , any ideas ?

Answer 6

@mukushla !!

Answer 7

Hello
what is \(Im(f)\) ?

Answer 8

Im is Image

Answer 9

never heard of it... what math u are in?

Answer 10

calculus

Answer 11

sorry not image, Im = imaginary

Answer 12

it must be image, same as range...it cant be imaginary because domain and range of \(f\) are made of real numbers according to problem statement

Answer 13

ok..

Answer 14

R -> R real numbers

Answer 15

i found the definition For z = x + iy means x Real part, y is the imaginary part of z. Notation: Re(z) = Im(z)

Answer 16

Re(z) = x, Im(z) = y. Let me see if I can dig up my old notes.

Answer 17

http://www.mediafire.com/download/ixj572fp5leh5lm/Complex_Variables_Notes_-_1.PDF

If anyone wants them for future reference, enjoy :)

Answer 18

considering \(Im\) as image of \(f\) for proving use the fact that \(\mathbb{Z}\) is a Discontinuous subset of \(\mathbb{R}\)

Answer 19

Can we just use the definition of continuity to show that it's can't jump from one integer value to the next, basically just what @mukushla said

Answer 20

What would the limit at infinity be?