Prove that √3+2√5is an irrational number
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Let us assume 3√2 + √5 = a rational number = p / q where p and q are integers and q is not 0. Let us assume that p / q is in reduced form and have no common factors and are prime to each other. ... Hence, 3 √2 + √5 is irrational.
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Let us suppose that 2√3+√5 is rational.
Let\: 2\sqrt{3}+\sqrt{5}=\frac{a}{b},\\where\:a,b\:are\: integers\\ \:and\:b≠0
Therefore, \:2\sqrt{3}=\frac{a}{b}-\sqrt{5}.
Squaring on both sides, we get
12=\frac{a^{2}}{b^{2}}+5-2\times \frac{a}{b}\times \sqrt{5}
Rearranging
\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}}{b^{2}}+5-12
\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}}{b^{2}}-7
\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}-7b^{2}}{b^{2}}
\implies\sqrt{5}=\frac{a^{2}-7b^{2}}{b^{2}}\times \frac{b}{2a}
After cancellation, we get
\implies\sqrt{5}=\frac{a^{2}-7b^{2}}{2ab}
Since , a ,b are integers , </p><p>\frac{a^{2}-7b^{2}}{2ab}
is rational,and so √5 is rational.
This contradicts the fact that √5 is irrational.
Hence , 2√3+√5 is irrational.
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