Question

Use strong induction to prove that$\sqrt{\mathbf{2}}$ is irrational. [Hint: Let$\mathit{P}\mathbf{\left(}\mathit{n}\mathbf{\right)}$ be the statement that$\sqrt{\mathbf{2}}\mathbf{=}\mathit{n}\mathbf{/}\mathit{b}$ for any positive integer b.]

Short answer

It has been proved that$\sqrt{2}$ is irrational.

Step-by-step solution

Define statement

As per hint, Let

$P\left(n\right):$There is no positive integer b such that $\sqrt{2}=\frac{n}{b}$.\

Use strong induction

The basis step: The statement is true for$n=1$ as$\sqrt{2}>1>\frac{1}{b}$ .

Inductive hypothesis: Assume that the statement$P\left(j\right)$ is true for all j with $j\le k$, where k is an arbitrary positive integer greater.

To show: The statement$P(k+1)$ is true.

Proof: By contradiction

Prove using contradiction

Let us assume that$\sqrt{2}=\frac{k+1}{b}$ for some positive integer b.

Squaring both sides,

$$\begin{gathered} 2=\frac{(k+1{)}^2}{b^2} \\ \Rightarrow 2b^2=(k+1{)}^2 \end{gathered}$$

This implies that$(k+1{)}^2$ is even.

Therefore, k+1 is even.

So$k+1=2t$ for some positive integer t.

Substituting this,

$$\begin{gathered} 2b^2=4t^2 \\ \Rightarrow b^2=2t^2 \end{gathered}$$

This implies b is even.

So b=2s for some positive integer s.

Thus

$$\begin{gathered} \sqrt{2}=\frac{k+1}{b} \\ =\frac{2t}{2s} \\ =\frac{t}{s} \end{gathered}$$

But$t\le k$ , so this contradicts the inductive hypothesis.

Hence our supposition is wrong.

Conclusion

Hence, there is no positive integer b such that $\sqrt{2}=\frac{n}{b}$.

Thus,$\sqrt{2}$ is irrational.