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Prove that the following are irrationals : $1. \frac{1}{\sqrt{2}}$ $2. 7\sqrt{5}$ $3. 6 + \sqrt{2}$
Prove that the following are irrationals :
- $\frac{1}{\sqrt{2}}$
- $7\sqrt{5}$
- $6 + \sqrt{2}$
This Question has 1 answers.
1. $\frac{1}{\sqrt{2}}$ is irrational. Assume $\frac{1}{\sqrt{2}} = \frac{p}{q}$, where $p$ and $q$ are integers. Then $\sqrt{2} = \frac{q}{p}$, which implies $\sqrt{2}$ is rational. But $\sqrt{2}$ is irrational, so the assumption is false, proving that $\frac{1}{\sqrt{2}}$ is irrational.
2. $1 + \sqrt{5}$ is irrational. Assume $1 + \sqrt{5} = \frac{p}{q}$, then $\sqrt{5} = \frac{p - q}{q}$, which implies $\sqrt{5}$ is rational. But $\sqrt{5}$ is irrational, so the assumption is false.
3. $6 + \sqrt{2}$ is irrational. Assume $6 + \sqrt{2} = \frac{p}{q}$, then $\sqrt{2} = \frac{p - 6q}{q}$, which implies $\sqrt{2}$ is rational. But $\sqrt{2}$ is irrational, so the assumption is false.
2. $1 + \sqrt{5}$ is irrational. Assume $1 + \sqrt{5} = \frac{p}{q}$, then $\sqrt{5} = \frac{p - q}{q}$, which implies $\sqrt{5}$ is rational. But $\sqrt{5}$ is irrational, so the assumption is false.
3. $6 + \sqrt{2}$ is irrational. Assume $6 + \sqrt{2} = \frac{p}{q}$, then $\sqrt{2} = \frac{p - 6q}{q}$, which implies $\sqrt{2}$ is rational. But $\sqrt{2}$ is irrational, so the assumption is false.
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