Hi,

I've seen a video that reminded me of the definition of continuity (that you can find here).
Somehow it seemed it was not what my "humble intuition" (sorry for that) excepted (although I remember it from school).. (no I don't confuse continuity with derivability ok).
My intuition is "a function that you can draw without lifting the pen"..
So I thought maybe I can find a function that is obviously not drawable without lifting the pen, but still match the classic definition for continuity, and I found a very simple one :
Take any real number x, write it in decimal, and then recompute it in base 8
So, in "math" (I try my best) term :

x = sum(d_n * 10^n]
f(x) = sum(d_n * 8^n]
(where d_n are the decimal digits
For instance :

0.5 ⇒ 5/8 = 0.625
0.1122 ⇒ 1/8 + 1/8^2 + 2/8^3 + 2 / 8^4 =0.14501953125

This function is defined for any real AND we can always find a box as little as we want around any point
for instance, we know that if
x is in [0.12364...(n digits)0 ;0.12364...(n digits)99999]

f(x) is in [f(0.12364...(n digits));f(0.12364...(n digits))+ 8 ^(-n)]
and we can take as many digits as we want.
And we also know that this function do have "jumps" everywhere, in any segment
Am I getting something wrong ?

Edited April 18Apr 18 by Edgard Neuman

Perhaps you can elaborate on how f(x) is continuous. It is clearly a discontinuous function to me.

5 hours ago, KJW said:

Perhaps you can elaborate on how f(x) is continuous. It is clearly a discontinuous function to me.

I call to your sagacity to figure it out by yourself 😉

22 minutes ago, Edgard Neuman said:

I call to your sagacity to figure it out by yourself 😉

When you refuse dialogue, it's only your loss, really 🤷‍♂️

3 minutes ago, KJW said:

When you refuse dialogue, it's only your loss, really 🤷‍♂️

ok, bye: !

no but seriously (sorry i'm drunk right now), if you take e (as in epsilon) is obvious that f([x-e;x+e]) is limited.. I mean there's a number of digits n so that [x-e;x+e] is in [(n digit of x)0;(n digit of x)99999....] and so f(x) is in (you know what)

I'll be much bold (and maybe wrong) : if there's a range limit on Y when there's a limit on X, THEN there's a limit on X when there's a limit on Y

1 hour ago, Edgard Neuman said:

sorry i'm drunk right now

That’s always so fun for everyone else!

11 hours ago, Edgard Neuman said:

ok, bye: !

no but seriously (sorry i'm drunk right now), if you take e (as in epsilon) is obvious that f([x-e;x+e]) is limited.. I mean there's a number of digits n so that [x-e;x+e] is in [(n digit of x)0;(n digit of x)99999....] and so f(x) is in (you know what)

I'll be much bold (and maybe wrong) : if there's a range limit on Y when there's a limit on X, THEN there's a limit on X when there's a limit on Y

I found my mistake. This function is discontinuous at every decimal number that ends with zeros.. and continuous otherwise

On 4/19/2025 at 10:54 AM, Edgard Neuman said:

I found my mistake. This function is discontinuous at every decimal number that ends with zeros.. and continuous otherwise

That's not the way of mathematics.

Prove that you were wrong!

Have some coffee first.