Question

Prove √3 irrational number​

Answer 1

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Step-by-step explanation:

Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1.

√3 = p/q

⇒ p = √3 q

By squaring both sides, we get,

p2 = 3q2  

p2 / 3 = q2 ------- (1)

(1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p2, then a divides p, where a is a positive integer)

Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p.

Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get,

(3c)2 / 3 = q2

9c2/3 =  q2  

3c2  =  q2  

c2  =  q2 /3 ------- (2)

Hence 3 is a factor of q (from 2)

Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is the contradiction to our assumption that p and q are co-primes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.

Answer 2

Answer:

Step-by-step explanation:

We have to prove that $\sqrt{3$ is an irrational number. Let us consider its opposite that $\sqrt{3$ is a rational number which means that

$\sqrt{3$ = $\frac{a}{b}$

where a and b are co-prime numbers and b$\neq$0.

By the formula that

Irrational number $\neq$ Rational number

Here $\sqrt{3}$ is an irrational number and $\frac{a}{b}$ is a rational number.

Hence, it is proved that $\sqrt{3}$ is an irrational number.