Question

prove that √5 is irrestional​

Answer 1

Step-by-step explanation:

Let's prove this by the method of contradiction-

Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.

⇒√5=p/q

⇒5=p²/q² {Squaring both the sides}

⇒5q²=p² (1)

⇒p² is a multiple of 5. {Euclid's Division Lemma}

⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}

⇒p=5m

⇒p²=25m² (2)

From equations (1) and (2), we get,

5q²=25m²

⇒q²=5m²

⇒q² is a multiple of 5. {Euclid's Division Lemma}

⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}

Hence, p,q have a common factor 5. this contradicts that they are co-primes. Therefore, p/q is not a rational number. This proves that √5 is an irrational number.

For the second query, as we've proved √5 irrational. Therefore 2-√5 is also irrational because difference of a rational and an irrational number is always an irrational number.

Answer 2

Answer:

Let us assume that √5 is rational.

Now, if it is a rational number it can be expressed as,

√5 = a/b ,where a and b are co-primes and b≠0.

Squaring on both sides we get,

⇒ (√5)² = (a/b)²

⇒ 5 = a²/b²

⇒ 5b² = a² ..............................(1)

⇒ b² = a²/5

now we get to know that 5 divides a².

⇒ 5 divides a...................................... (2)

Now, a = 5c

Squaring on both sides,

⇒ a² = 25c²

From (1),

⇒ 5b² = 25c²

⇒ b² = 5c²

⇒ b²/5 = c²

⇒ 5 divides b² which also further means that 5 divides b.

Hence a and b have 5 as a common factor.

Our assumption was wrong.

Therefore a and b are not co-primes, which means that our assumption was wrong and √5 is irrational.