Math, asked by sreenidhireddy1811, 8 months ago

prove that √5 is an irrational number

Answers

Answered by MysteriousAryan
8

Answer:

\huge{\mathcal{\underline{\red{AnSWer}}}}

Given: √5

We need to prove that √5 is irrational

✍︎✍︎Proof:

Let us assume that √5 is a rational number.

Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒√5=p/q

On squaring both the sides we get,

⇒5=p²/q²

⇒5q²=p² —————–(i)

p²/5= q²

So 5 divides p

p is a multiple of 5

⇒p=5m

⇒p²=25m² ————-(ii)

From equations (i) and (ii), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5

⇒q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number

Hence proved

✍︎✍︎✍︎Some more info ✍︎✍︎✍︎

  • A rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero,
  • An irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating.

Answered by singhkarishma882
5

\huge\red{\boxed{\mathfrak{☆AnSwEr☆}}}

\sf\green {\underline {\sf To\:Prove,}}

\sqrt{5} is irrational.

\sf\orange {\underline {\sf Let's\:assume,}}

\sqrt{5} as a rational number.

We can find integers a and b such \sqrt{3} = \frac{a}{b} .

Suppose a and b have common factors other than 1, then we can divide by the common factor, and assume that a and b are co-primes.

So, b\sqrt {3}\:=\:a

Squaring on both the sides, we get :-

 {(b \sqrt{3} )}^{2}  =  {(a)}^{2}

 {3b}^{2}  =  {a}^{2}

 {b}^{2}  =  { \frac{a}{3} }^{2}

a = 3c

By substituting for a, we get :-

a = 3c

 {3b}^{2}  =  {(3c)}^{2}

 {3b}^{2}  =  {9c}^{2}

 {b}^{2}  =  { \frac{9c}{3} }^{2}

 {b}^{2}  =  {3c}^{2}

Since,

a and b have at least 3 as a common factor.

But,

this contradicts the fact that a and b have no common factors other than 1.

Reason,

this contradiction arised because of our wrong assumption that \sqrt {3} is rational.

So we conclude that \sqrt{3} is irrational.

\huge\mathbb\color {grey}Might\:Help\:You

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