Show $|u^n| = n|u|$ for all strings $u$ and all $n$

Question

Can anyone please help me with this homework question on automata from Peter Linz?

Use induction on $n$ to show that $|u^n| = n|u|$ for all strings $u$ and all $n$.

Answer 1

First we need to know what the cardinality of a string is. For example what is the cardinality of the string 'xyz'. I would guess it's 3 in this case (the number of characters in the string), but this should be confirmed.

Next we need to know how $u^n$ is defined. Again, it might be helpful to consider the example 'xyz'. I would guess $(xyz)^2$ would be xyzxyz, so $u^n$ is n lots of the string u concatenated together into one string.

Based on these assumptions, the proof by induction would be constructed in this way:

Base case $(n=1):$ $|u^1| = |u| = 1|u|$

Inductive step:

Suppose $|u^k|=k|u|$

Then $|u^{k+1}| = |u^k| + |u|$ (by definitions of cardinality of string and $u^n$)

= k|u| +|u| (employing the inductive hypothesis)

= (k+1)|u| (by definition of cardinality of string)

The result follows by induction

I'm not sure if $n = 0$ needs to be considered. If there is a definiton of $u^0$ it woild presumably be $u^0 =$ the empty string $(\lambda)$, the cardinality of which I would presume to be $0$, so the base case would then be

Base case $(n=0):$ $|u^0| = |\lambda| = 0 = 0|u|$ so it would still work and the rest of the proof would be identical.