French, asked by kfkfkkdkdjd, 1 month ago

Prove that √3-√5 is irrational

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Answered by Fαírү
102

To Prove :

 \sf{ \sqrt{3} -  \sqrt{5}  \:  \: is \: irrational}

Let us assume that  \sqrt{3}  -  \sqrt{5} is rational, which let us take as a.

Now,

 \sf{ \sqrt{3}  -  \sqrt{5}  = a}

Squaring on both sides,

 \sf{( \sqrt{3}  -  \sqrt{5} )^{2}  =  {a}^{2} }

 \sf{8 - 2 \sqrt{15}  =  {a}^{2} }

 \sf{ - 2 \sqrt{15}  =  {a}^{2}  - 8}

 \sf{ \sqrt{15}  =  \frac{ {a}^{2}  - 8}{ - 2} }

 \sf{ \frac{ {a}^{2}  - 8}{ - 2}} is a rational number, whereas  \sqrt{15} is irrational.

We know that, One rational number cannot be equal to an irrational number.

Which means that,

Our assumption was wrong.

 \sqrt{3}  -  \sqrt{5} is irrational.

 \sf{ \red{Hence \: Proved!}}

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