Math, asked by vanamvenkateshwarlu8, 4 months ago

prove that √3+√5is an irrational number​

Answers

Answered by ak7591422
2

Answer:

let as assume

 \sqrt{3 +  \sqrt{5} }

is a rational number. we can find the co prime integer

so that

 \sqrt{3 +  \sqrt{5} }

 \sqrt{3 +  \sqrt{5 = a \div b } }

 \sqrt{5 = a \div b -  \sqrt{3} }

 \sqrt{5 =  \sqrt{3 + a \div b} }

therefore

 \sqrt{3 + a \div b}

is a rational , so

 \sqrt{5}

is rational

But this contradict the fact that

 \sqrt{5}

is irrational number

hence,

 \sqrt{3 +  \sqrt{5} }

is irrational number

Answered by Anonymous
4

Step-by-step explanation:

Let √3+√5 be a rational number. A rational number can be written in the form of p/q where p,q are integers. p,q are integers then (p²+2q²)/2pq is a rational number. ... Therefore, √3+√5 is an irrational number.

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