Prove that √p + √q is irrational, where p, q are primes

Solution:

Consider that √p + √q is rational.

Now √p + √q = a , where a is rational.

Therefore, √q = a - √p

By squaring on both sides, we get q = a² + p - 2 a√p

Using the algebraic identity

(a - b)² = a² + b² - 2ab

√p = a² + p - q/2a, which is a contradiction as the right hand side is a rational number while √p is irrational, since p is a prime number

Therefore, √p + √q is irrational

✦ Try This: Prove that √5 is irrational

Let us assume that √5 is a rational number

So it can be expressed in the form p/q where p,q are co-prime integers and q≠0

√5 = p/q.

On squaring both the sides we get,

5 = p²/q²

5q² = p² --- (i)

p²/5 = q²

So 5 divides p

p is a multiple of 5

p = 5m

p² = 25m² --- (ii)

From equations (i) and (ii), we get,

5q² = 25m²

q² = 5m²

q² is a multiple of 5

q is a multiple of 5.

p, q have a common factor 5. This contradicts our assumption that they are co-primes.

p/q is not a rational number

Therefore, √5 is an irrational number

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 1

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NCERT Exemplar Class 10 Maths Exercise 1.3 Problem 14

Prove that √p + √q is irrational, where p, q are primes

Summary:

Irrational numbers are those real numbers that cannot be represented in the form of a ratio. In other words, those real numbers that are not rational numbers are known as irrational numbers. √p + √q is irrational, where p, q are primes. Hence proved

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☛ Related Questions:

• Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer . . . .
• Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, . . . .
• Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive int . . . .