Answer:
Given, \[\sqrt{2}\] is irrational number. Let \[\sqrt{2}=m\] Suppose, \[5+3\text{ }\sqrt{2}\] is a rational number. So, \[5+3\text{ }\sqrt{2}=\frac{a}{b}\] \[(a\ne b,b\ne 0)\] \[3\sqrt{2}=\frac{a}{b}-5\] \[3\sqrt{2}=\frac{a-5b}{b}\] or \[\sqrt{2}=\frac{a-5b}{3b}\] So, \[\frac{a-5b}{3b}=m\] But \[\frac{a-5b}{3b}\] is rational number, so m is rational number which contradicts the fact that \[m=\sqrt{2}\] is irrational number. So, our supposition is wrong. Hence, \[5+3\sqrt{2}\] is also irrational.
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