Question

prove that √3 is irrational​

Answer 1

Answer:

Let us assume to the contrary that √3 is a rational number.

It can be expressed in the form of p/q

where p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p2/q2 (Squaring on both the sides)

⇒ 3q2 = p2………………………………..(1)

It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

So we have p = 3r

where r is some integer.

⇒ p2 = 9r2………………………………..(2)

from equation (1) and (2)

⇒ 3q2 = 9r2

⇒ q2 = 3r2

Where q2 is multiply of 3 and also q is multiple of 3.

Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.

Step-by-step explanation:

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Answer 2

✌✌Hii Good morning mate Here is your Answer✌✌

Let us assume to the contrary that √3 is a rational number.

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/q

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3r

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)⇒ 3q2 = 9r2

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)⇒ 3q2 = 9r2⇒ q2 = 3r2

Let us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)⇒ 3q2 = 9r2⇒ q2 = 3r2We have two cases to consider now.

CASE 1

Suppose that r is even. Then r2 is even, and 3r2 is even which implies that q2 is even and so q is even, but this cannot happen. If both q and r are even then gcd(q,r)≥2 which is a contradiction.

CASE 2

Now suppose that r is odd. Then r2 is odd and 3r2 is odd which implies that q2 is odd and so q is odd. Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N.

Therefore:-

• q2=3r2

• (2m−1)2=3(2n−1)2

• 4m2−4m+1=3(4n2−4n+1)

• 4m2−4m+1=12n2−12n+3

• 4m2−4m=12n2−12n+2

• 2m2−2m=6n2−6n+1

• 2(m2−m)=2(3n2−3n)+1

We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3.

Hence the root of 3 is an irrational number.

✌✌Hii. Finally we prooved this✌✌

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