Math, asked by ayaanahmed0711, 4 months ago

Prove that √3+√5 is irrational​

Answers

Answered by XxShAnTaNuxX
0

\huge \underline \frak \pink {★彡A᭄ɴsᴡᴇʀ彡★}

→ √3 + √5

→ 1.732... + 2.236...

→ 3.961... [Non - Terminating nD Non - Repeating ]

\large\colorbox{orange}{May Be Helpful ✌️ Dear ✌️}

Answered by kailashmannem
50

 \Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}

  • √3 + √5

 \Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: prove:-}}}}}}}

  • It is a irrational number.

 \Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Proof:-}}}}}}}

  • √3 + √5

  • First,

Let us assume that √3 + √5 is an rational number.

  • √3 + √5 =  \sf \dfrac{a}{b}

  • where,

  • a and b are co - primes and b  \sf \neq 0.

  • Now,

 \sf \sqrt{5} \: = \: \dfrac{a}{b} \: - \: \sqrt{3}

  • Squarring on both sides,

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

 \sf 5 \: = \: \dfrac{a^{2}}{b^{2}} \: + \: 3 \: - \: 2 . \dfrac{a}{b} . \sqrt{3}

 \sf 5 \: - \: 3 \: = \: \dfrac{a^{2}}{b^{2}} \: - \: 2 . \dfrac{a}{b} . \sqrt{3}

 \sf 2 \: = \: \dfrac{a^{2}}{b^{2}} \: - \: 2 . \dfrac{a}{b} . \sqrt{3}

 \sf 2 \: = \: \dfrac{a^{2}}{b^{2}} \: \dfrac{- \: 2a}{b} . \sqrt{3}

 \sf \dfrac{2a}{b} . \sqrt{3} \: + \: 2 \: = \: \dfrac{a^{2}}{b^{2}}

 \sf \dfrac{2a}{b} . \sqrt{3} \: = \: \dfrac{a^{2}}{b^{2}} \: - \: 2

  • Taking LCM on RHS,

 \sf \dfrac{2a}{b} . \sqrt{3} \: = \: \dfrac{a^{2} \: - \: 2b^{2}}{b^{2}}

 \sf \sqrt{3} \: = \: \dfrac{a^{2} \: - \: 2b^{2}}{b^{2}} \: * \: \dfrac{b}{2a}

 \sf \sqrt{3} \: = \: \dfrac{a^{2} \: - \: 2b^{2}}{\cancel{b^{2}}} \: * \: \dfrac{\cancel{b}}{2a}

 \sf \sqrt{3} \: = \: \dfrac{a^{2} \: - \: 2b^{2}}{b} \: * \: \dfrac{1}{2a}

 \boxed{\pink{\sf \sqrt{3} \: = \: \dfrac{a^{2} \: - \: 2b^{2}}{2ab}}}

  • Here, a and b are Integers.

Therefore,

  •  \sf \dfrac{a^{2} \: - \: 2b^{2}}{2ab} is a rational number.

  • But, this contradicts the fact that √3 + √5 is a irrational number.

Therefore,

  • Our assumption is wrong.

  •  \underline{\boxed{\blue{\therefore{\sf \sqrt{3} \: + \: \sqrt{5} \: is \: an \: irrational \: number.}}}}
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