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  • Prove that Squaeroot 3+ Squareroot 5 is irrational

Prove that \sqrt{3}+\sqrt{5} is irrational.

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We will do it by the method of contradiction: We will assume \sqrt{3}+\sqrt{5} is a rational number. If it leads to some absurd outcome then it is a wrong assumption.

Let \sqrt{3}+\sqrt{5} is a rational number
  \therefore \sqrt{3}+\sqrt{5}= \frac{a}{b},b\neq 0,a,b\, \epsilon z
  \sqrt{5}= \frac{a}{b}-\sqrt{3}
Squaring both side.
\left ( \sqrt{5} \right )^{2}= \left ( \frac{a}{b} -\sqrt{3}\right )^{2}
5= \left ( \frac{a}{b} \right )^{2}+\left ( \sqrt{3} \right )^{2}-2\left ( \frac{a}{b} \right )\left ( \sqrt{3} \right )
 (Using (a – b)2 = a2 + b2 = 2ab)

   \left ( \frac{a}{b} \right )^{2}+3-2\left ( \frac{a}{b} \right )\left ( \sqrt{3} \right )= 5
     \frac{a^{2}}{b^{2}}+3-5= 2\sqrt{3}\left ( \frac{a}{b} \right )
  \frac{a^{2}-2b^{2}}{b^{2}}= 2\sqrt{3}\left ( \frac{a}{b} \right )
  \frac{b\left ( a^{2} -2b^{2}\right )}{2ab^{2}}= \sqrt{3}

    \frac{a^{2}-2b^{2}}{2ab}= \sqrt{3}
LHS is a rational number but RHS is an irrational number.
This is not possible, hence our assumption was wrong.
Hence \sqrt{3}+\sqrt{5} is a irrational number.

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