Question
Short-Answer Questions:
Show that $\frac{\sqrt2}{3}$ is irrational.
Show that $\frac{\sqrt2}{3}$ is irrational.
✓
Answer
If possible, let $\frac{\sqrt2}{3}$ be rational.
$\Rightarrow\frac{1}{3}\sqrt2$ is rational.
Now, 3 is rational, $\frac{1}{3}\sqrt2$ is rational.
$\Rightarrow\Big(3\times\frac{1}{3}\sqrt2\Big)$ is rational.
$\Rightarrow\sqrt2$ is rational.
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\frac{\sqrt2}{3}$ is rational.
Hence, $\frac{\sqrt2}{3}$ is irrational.
$\Rightarrow\frac{1}{3}\sqrt2$ is rational.
Now, 3 is rational, $\frac{1}{3}\sqrt2$ is rational.
$\Rightarrow\Big(3\times\frac{1}{3}\sqrt2\Big)$ is rational.
$\Rightarrow\sqrt2$ is rational.
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\frac{\sqrt2}{3}$ is rational.
Hence, $\frac{\sqrt2}{3}$ is irrational.
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