Question

for numbers from 0 to 9, we can construct (10)^n numbers of n digits.
show that we can construct infinite numbers for unlimited "n".

Answer 1

Ermm... in same way show that N is infinite (but wouldn't this be with a definition ?"

Answer 2

no its not

Answer 3

0-9^n

Answer 4

o through 9 to the n power

Answer 5

Pardon @usercode3rror ?

Answer 6

@ikram002p i think it wants you to show i diagram of the function but idk it rly looks like a def but im not sure i think its 0^n 1^n 2^n 3^n 4^n 5^n 6^n 7^n 8^n 9^n but its inf many so idk

Answer 7

Not sure I understand the question. It sounds like you're just asked to prove \(\lim\limits_{n\to\infty}n^n=\infty\)?

Answer 8

induction?

Answer 9

hmmm @siths well know... it proves that natural number set is infinite.
@unklerhaukus don't we need that fact to show induction ? i thought of well ordering principle at first place :O.