prove that √3 is irrational
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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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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:-
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.
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