English, asked by Anonymous, 5 months ago

Write a short story in which an ancient trunk, a pianist and a school boy play a
pivotal role.
\huge\underline\mathbb\pink{No/: Spam}
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Answers

Answered by Anonymous
9

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

Let us assume to the contrary that √5 is a rational number.

is a rational number.It can be expressed in the form of p/q

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

is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √5= p/q

= p/q⇒ 5 = p2/q2 (Squaring on both the sides)

= p2/q2 (Squaring on both the sides)⇒ 5q2 = p2………………………………..(1)

q2 = p2………………………………..(1)It means that 5divides p2 and also 5divides p because each factor should appear two times for the square to exist.

divides p because each factor should appear two times for the square to exist.So we have p = 5r

rwhere r is some integer.

rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)

rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)

rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)⇒ 5q2 = 25r2

r2⇒ q2 = 3r2

r2⇒ q2 = 3r2Where q2 is multiply of 5 and also q is multiple of 5.

.Then p, q have a common factor of 5. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √5 is an irrational number

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