Prove that root 5 + root 3 is an irrational number

1.	Prove that root 5 + root 3 is an irrational number
Answer» If possible, let\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}\xa0be a rational number equal to x. Then,x =\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}{tex}\\Rightarrow{/tex}\xa0x2 = ({tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex})2\xa0{tex}\\Rightarrow{/tex}\xa0x2 = ({tex}\\sqrt{3}{/tex})2 + ({tex}\\sqrt{5}{/tex})2 + 2\xa0{tex}\\times{/tex}\xa0{tex}\\sqrt{3}{/tex}\xa0{tex}\\times{/tex}\xa0{tex}\\sqrt{5}{/tex}= 3 + 5 + 2{tex}\\sqrt{15}{/tex}= 8 + 2{tex}\\sqrt{15}{/tex}{tex}\\Rightarrow{/tex}\xa0x2 - 8 = 2{tex}\\sqrt{15}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{x^2 -8}{2}{/tex}\xa0=\xa0{tex}\\sqrt{15}{/tex}Now, x is rational{tex}\\Rightarrow{/tex}\xa0x2 is rational{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{x^2 - 8}{2}{/tex}\xa0is rational{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt{15}{/tex}\xa0is rationalBut,\xa0{tex}\\sqrt{15}{/tex}\xa0is irrational.Thus, we arrive at a contradiction. So, our supposition that\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}\xa0is rational is wrong.Hence,\xa0{tex}\\sqrt{3}{/tex}\xa0+\xa0{tex}\\sqrt{5}{/tex}\xa0is an irrational number.

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