Question
Show that $\big(4+3\sqrt2\big)$ is irrational.
✓
Answer
If possible, let $\big(4+3\sqrt2\big)$ be rational.
Then 4 and $3\sqrt2$ are rational.
$\Rightarrow4+3\sqrt2-4$ is rational $\dots(\because$ diference of two rationals is rational$)$
$\Rightarrow3\sqrt2$ is rational
$\Rightarrow\sqrt2$ is rational $\dots(\because$ 3 is rational$)$
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\big(4+3\sqrt2\big)$ is rational.
Hence, $\big(4+3\sqrt2\big)$ is irrational.
Then 4 and $3\sqrt2$ are rational.
$\Rightarrow4+3\sqrt2-4$ is rational $\dots(\because$ diference of two rationals is rational$)$
$\Rightarrow3\sqrt2$ is rational
$\Rightarrow\sqrt2$ is rational $\dots(\because$ 3 is rational$)$
This contradicts the fact that $\sqrt2$ is irrational.
The contradiction arises by assuming that $\big(4+3\sqrt2\big)$ is rational.
Hence, $\big(4+3\sqrt2\big)$ is irrational.
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