Math, asked by Anonymous, 8 months ago

/huge/bold/pink/underline{Question}

Prove that √5 is irrational??​

Answers

Answered by koonergavan
3

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

Answered by silentloffer
3

\Huge{\bold{\mathrm{\underline{\red{ANSWER}}}}}

Let √5 be a rational number 

Therefore, √5= p/q  [ p and q are in their least terms i.e., HCF of (p,q)=1 and q ≠ 0

On squaring both sides, we get 

                   p²= 5q²                                                                                    ...(1)

Clearly, 5 is a factor of 5q²

⇒ 5 is a factor of p²                                                                    [since, 5q²=p²]

⇒ 5 is a factor of p

 Let p =5 m for all m ( where  m is a positive integer)

Squaring both sides, we get 

            p²= 25 m²                                                                                          ...(2)

From (1) and (2), we get 

           5q² = 25m²      ⇒      q²= 5m²

Clearly, 5 is a factor of 5m²

⇒       5 is a factor of q²                                                      [since, q² = 5m²]

⇒       5 is a factor of q 

Thus, we see that both p and q have common factor 5 which is a contradiction that H.C.F. of (p,q)= 1

     Therefore, Our supposition is wrong

Hence √5 is not a rational number i.e., irrational number.

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