Question

prove that √5 is irrational​

Answer 1

Here we go with the proof; Let us assume that √5 is rational So, √5 is a rational number All rational numbers are in p/q form So, √5 = p/q , where p≠0 and p&q are integers Squaring on both sides (√5)²=(p/q)² 5=p²/q² 5q²=p²–(1) 5 divides p [p is any prime number and a is any positive integer and p divides a² then p divides a] So, p is a positive integer—(2) Then, p=5r, ...

Answer 2

Answer:

let's √5 be rational and let's it's simplest a|b

then a and b are integre having no common factor others then 1 and b is not equal 0.

now,

√5=a/b =5=A2/b2. (on square both side)

= 5b2 = A2

=5 divides A2( 5 divides 5b2)

=5 divides a

(5 is a prime and 5 divides b2 =5

divides A)

let's a= 5c for some integers c

putting a=5c in 1. We get

5b2 = 25c2=b2=5c2

5 divides b2. ( 5divides 5c2}

5divides b

( 5is a prime numbers and 5divides b2

5 divides b)

thus 5 is a common factor of a and b.

but this contradict fact a and b have no common factor others than 1.

Step-by-step explanation:

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