10th Class Mathematics Solved Paper - Mathematics-2018

  • question_answer 7) Given that \[\sqrt{2}\] is irrational, prove that \[\left( 5+3\sqrt{2} \right)\]is an irrational number.

    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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