Proofs by Induction: Divisibility

Question

In one the problems on divisibility we needed to show that $10k−(−1)k$ is always divisible by $11$. Sketch a proof by induction for this.

How would one start this process and how does one become good at proofs?

Answer 1

We answer the second question first: practice, practice, practice.

The main question is undoubtedly intended to be about $10^k-(-1)^k$.

The result is clearly true for $k=1$. We now show that if it is true at $j$, it is true at $j+1$. We have $$10^{j+1}-(-1)^{j+1}=11(10^j)-10^j-(-1)^{j+1}=11(10^j)-(10^j-(-1)^j).$$ This should let you complete the induction step.

Remark: Congruence notation makes for a much more natural proof. Since $10\equiv -1\pmod{11}$, it follows that $10^k\equiv (-1)^k\pmod{11}$.