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I understand induction with one variable well, however I am not sure what to do when there are 2 or more variables.

The problem I came across is following:

Prove that $a^r \ge 1$, where $r \in \mathbb{N}$ and $a \in \mathbb{R} \wedge a \ge 1$

My solution which I am not sure whether is right:

1) For $r=0$:

$a^0 = 1$, $1\ge1$

2) Now assuming $a^r \ge 1$, I try to prove that $a^{r+1} \ge 1$.

$a^{r+1} = a^r * a$

As assumed $a^r \ge 1$ and also $a \ge 1$, therefore $a^r * a \ge 1$.


Is this proof correct or do I have to do it for $a$ somehow as well?

John Smith
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1 Answers1

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Your proof looks fine to me. For this proof, there's no need to induct on $a$.

The induction proof as is assumes that $a$ is just some arbitrary number for which $a \in R \wedge a \geq 1$ is true. This is enough to show that any $a$ that fulfills $a \in R \wedge a \geq 1$ will work (since you can freely substitute any $x$ such that $x \in R \wedge x \geq 1$ and the proof doesn't change).

Dennis Meng
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