Question

Is this function continuous? Justify

Answer 1

\(\large \mathbb{f : Z \rightarrow Z}\) where \(f(x) = \lfloor x\rfloor\)

Answer 2

:D yep

Answer 3

thanks! could you please explain why

Answer 4

2 ways i knw to show
1- its discrete function with 1-1 relation
2- for any x in Z there is n s.t , x in (x-n, x+n) for any f(x) in ( f(l) ,f(g) ) in this case it would be f(x) in ( f(x-n) ,f(x+n) )

Answer 5

its a greatest integer function right?

Answer 6

floor function i guess :3

Answer 7

yes :)

ikram -
are all discrete functions continuous ?

i see you're using limit definition in second method, im still going through it..

Answer 8

yes
then
\[if~~f(x)=[x]\\then\\if,[x]=n \rightarrow n \le x \le n+1\]

Answer 9

first question ... only on Z , if all x has f(x) , and the function is monotone then continues .
second question ... its neighborhood def thingy :3

Answer 10

yesh :3

Answer 11

yep , some how

Answer 12

Okay aren't you using below definition in second method ?

\(\rm f(x)\) is continuous if
\[\rm \forall \epsilon \gt 0, \exists \delta \gt 0 \;s.t \\~\\\; |x-a|\lt \delta \implies |f(x)-f(a)| \lt \epsilon\]

Answer 13

ok some basic properties if u wanna know--\[1.[x+n]=n+[x],x \in Z\\2.[-x]=-[x], x \in Z\\3.[-x]=-[x]-1,x \notin Z\\4.[x]-[-x]=2n, x<n, n \in Z\\5.[x]-[-x]=2n+1, x=n+1,n \in Z\\6.[x] \ge n \rightarrow x \ge n\\7.[x+y] \ge [x]+[y]\]

Answer 14

yes they are, choose delta = 1/2 and you are done.

Answer 15

let \(\epsilon\) be given, choose \(\delta = 1/2\) then \(|x-a|< 1/2 \implies x=a \implies f(x) = f(a)\implies |f(x)-f(a)|=0<\epsilon\)

Answer 16

oh im convinced its continuous based on continuity definition, but it looks all broken on graph lol

Answer 17

why do you guys keep using the floor function... the domain is the integers

for any integer \(z\) we have\[ z=\lfloor z\rfloor\]

Answer 18

so the above is really f(x)=x

Answer 19

dont u use graph , ur domain is Z, now if u have domain of C how would u also confenss yourself with graphing too ;)

Answer 20

here is how u can understand it by graphing

Answer 21

I get it. domain is integers, so we only have integers to approach in domain and in codomain

Answer 22

exactly :)

Answer 23

If it was defined in \(\rm f:R\rightarrow R\), it wont be continuous because the limit wont exist at each integer. Got it completely thanks !

Answer 24

ovs :P

Answer 25

or R -> Z

same thing really as the image is integral

Any function on any subset of Z is continuous.

Answer 26

Ohh does that mean jump discontinuity is not a real discontinuity as per the definitoin ?

Answer 27

x mod n is not continues :)

Answer 28

R->R

f(x) = 1/x

the function itself is not defined at x=0, so the domain wont include 0. that means the function is continuous right ?

Answer 29

yep

Answer 30

Let \(f:D\rightarrow \mathbb{R}\), where \(D\subseteq Z\).
Let \(a\in D\), Let \(\epsilon > 0\), choose \(\delta = \frac{1}{2}\), then for all \(x\in D\).
\(|x-a|< \delta \implies |x-a|<\frac{1}{2} \implies \\ x=a \implies f(x)=f(a) \implies |f(x)-f(a)|=0<\epsilon\)

Answer 31

\(\rm f:D\rightarrow R, ~f(x) = \lfloor x \rfloor\) that works for discrete functions because you're only interested in\(\mathbb {a \in D \subset Z}\), so any \(\delta \gt 0\) works for any given \(\epsilon \ge 0\) as this function is just all points.. idk how to say

Answer 32

i got what ur saying :)

Answer 33

the function is not 'jumping' in its domain of Z

Answer 34

its jumping with respect to R