Math, asked by aryankapoor47, 10 months ago

Prove that 5 - √3 is irrational, given that √3 is irrational​

Answers

Answered by AdorableAstronaut
40

 \huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}

  • To Prove : 5 -  \sqrt{3}is irrational.

  • Given :  \sqrt{3} is already rational.

Let us assume that 5 -  \sqrt{3} is rational.

So now,

5 -  \sqrt{3} =  \frac{a}{b} ( Where  \frac{a}{b} is a rational number )

Taking 5 to the other side,

 \sqrt{3} \:  =  \frac{a}{b} - 5

Taking the LCM,

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

We know that,   \frac{a - 5b}{b} is a rational number. But \sqrt{3}\: is irrational, which is given.

So our assumption is wrong, and 5 -  \sqrt{3} is an irrational number.

 \sf{Hence \: Proved! \: }

Answered by Anonymous
71

 \huge{ \bold{ \underline{ \blue{ \sf{Detailed \: Answer}}}}}

we have to prove that 5-√3 is irrational.

let us assume that 5-√3 is irrational.

we know that if any no is rational then it can be written in the form p/q

Hence,

5 -  \sqrt{3}  \: can \: be \: written \: in \: the \: form \:  \frac{a}{b}

where a,b(b≠0 ) are co- prime .

thus ,

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

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

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

 \sqrt{3}  =  - ( \frac{a - 5b}{b} )

Here ,

 \frac{ - a + 5b}{b}  \: is \: rational

But √3 is irrational.

since, Rational≠iraational

this is a contradiction

∴ our assumption is wrong

Hence , 5-√3 is irrational.

\huge{\bold{ Hence\: proved }}

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