Question

Prove that (10) 3 is an irrational number.

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Answer 1

Step-by-step explanation:

Atmospheric pressure, also known as barometric pressure, is the pressure within the atmosphere of Earth. The standard atmosphere is a unit of pressure defined as 101,325 Pa, which is equivalent to 760 mm Hg, 29.9212

Answer 2

Answer:

Question :

Prove that sqaure root of 3 is an irrational number.

Solution :

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

It can be expressed in the form of p/q

It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.

⇒ √3 = p/q

⇒ 3 = p^2/q^2 (Squaring on both the sides)

⇒ 3q^2 = p^2 ------------------------(1)

It means that 3 divides p^2 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.

⇒ p^2 = 9r^2 -----------------------(2)

from equation (1) and (2)

⇒ 3q^2 = 9r^2

⇒ q^2 = 3r^2

We have two cases to consider now.

Case I

Suppose that r is even. Then r^2 is even, and 3r^2 is even which implies that q^2 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 II

Now suppose that r is odd. Then r^2 is odd and 3r^2 is odd which implies that q^2 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

q^2=3r^2

(2m−1)2=3(2n−1)^2

4m^2−4m+1=3(4n2−4n+1)

4m^2−4m+1=12n^2−12n+3

4m^2−4m=12n^2−12n+2

2m^2−2m=6n^2−6n+1

2(m^2−m)=2(3n^2−3n)+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 r^2=3.

Hence the root of 3 is an irrational number.

$\underline \bold \red{Hence \:Proved}$