Question

proove that √5 is an irrational no​

Answer 1

Answer

ANSWER

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

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Here is answer :

Let us assume on the contrary that √5 is a rational number.Then, there exist co-prime positive integers 'a' and 'b' such that

$\sqrt{5} = \frac{a}{b}$

Square on both sides..

$\sqrt{ {5} }^{2} = \frac{ {a} \: \: ^{2} }{b}$

=> 5b² = a²

=> 5 | a²

=> 5 | a [according to the theorem]. ........(1)

=> a = 5c for some positive integer c

=> a² = 25c² (square on both sides)

=> 5b² = 25c². [ a² = 5b²]

=> b² = 5c²

=> 5 | b²

=> 5 | b [according to the theorem].........(2)

From eqⁿ (1) & (2)

This contradicts the fact that 'a' and 'b' are co-prime.

Hence, √5 is an irrational number

Proved