Question

Prove √3 in irrational by using that prove 5-√3 is irrational

Answer 1

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${\huge {\underline {\underline {\mathfrak {\pink{QuEsTiOn}}}}}}$

Prove √3 in irrational by using that prove 5-√3 is irrational

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${\huge {\underline {\underline {\mathfrak {\pink{AnSwEr}}}}}}$

Let us assume that 5 -√3 is rational.

Let ,

 5 - √3 = r , where "r" is rational

5 - r = √3

Here,

LHS is purely rational.

But on the other hand ,RHS is irrational.

This leads to a contradiction.

Hence,5-√3 is irrational

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Answer 2

$\huge\underline{\overline{\mid{\bold{\pink{QueStioN-}}\mid}}}$

Prove √3 in irrational by using that prove 5-√3 is irrational

$\mathfrak{\huge{\orange{\underline{\underline{Answer :}}}}}$

$\Large{\underline{\underline{\bf{SoLuTion:-}}}}$

To prove that √3 is an irrational number,We have to find the square root of √3 by Long Division Method.

★ √3 = 1.7302050807

We observe that the decimal representation of √3 is neither terminating nor repeating.

$\dag \: \sf \red{Hence \: \sqrt{3 } \: is \: an \: irrational \: number}$

We shall prove this by the method of contradiction. If possible,let us assume that √3 is a rational number. Then,

√3 = $\sf\dfrac{p}{q}$ (Here, p and q are integers having no common factor and q ≠ 0)

→ 3 = $\sf \dfrac{ {p}^{2} }{ {q}^{2} }$ (Squaring both sides)

→ p² = 3q².... i)

→ q² is an even integer

→ p = 3m,where m is an integer

→ 3q² = 9m² (Using equation i)

→ q² = 3m²

→ q² is an even integer

→ q is an even integer

So, both p and q are even integers and therefore have a common factor 3. But, this contradicts that p and q haven't any common factor. Thus, our assumption is wrong

$\dag \: \sf \red{Hence \: \sqrt{3 } \: is \: an \: irrational \: number} \:$

Now ,

$\huge\underline{\overline{\mid{\bold{\pink{To Prove:-}}\mid}}}$

5-√3 is irrational

Let us assume the given number be rational and we will write the given number in p/q form

$5 - \sqrt{3 =}$

$= p \div q$

$\sqrt{3 = }$

$= 5q - p \div q$

We observe that LHS is irrational and RHS is rational, which is not possible.

This is contradiction.

Hence our assumption that given number is rational is false

⇒

$5 - \sqrt{3}$

is irrational.

Hence,

$\Huge\bold{\underline{\underline{\green{ Proved}}}}$