Proving Irrationality of Numbers

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In this section, we will discuss proving irrationality of numbers.We will prove √2, √3 and √2/√5 etc. are irrational numbers using Fundamental Theorem of Arithmetic.

In proving irrationality of these numbers, we will use the result that if a prime p divides a ² then it divides ‘a’ also. We will prove the irrationality of numbers by using the method of contradiction.

Examples :
1) Prove that √2 is an irrational number.

Solution :
Let us assume that √2 is a rational number. So,
√2 = a/b (where a and b are prime numbers with HCF = 1)
Squaring both sides
2 = a ² /b ²
⇒ 2b ² = a ²
⇒ 2 | a ²
⇒ 2 | a ------------> (1) [by Theorem -> Let p be a prime number. If p divides a ² , then p divides a, where a is a positive integer]
a = 2c for some integer c.
Squaring both sides
a ² = 4c ²
2b ² = 4c ² [ Since a ² =2b ² ]
⇒ b ² = 2c ²
⇒ 2 | b -------------> (2)
∴ From (1) and (2) we obtain that 2 is a common factor of ‘a’ and ‘b’. But this contradicts the fact.
So our assumption of √2 is a rational number is wrong.
Hence, √2 is an irrational number.
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2) √5 + √3 is an irrational number.
Solution :
Let √5 + √3 is a rational number equal to a/b.
√5 + √3 = a/b
√5 = a/b - √3
Squaring both sides
5 = (a/b - √3) ²
⇒ 5 = a ² /b ² - 2a√3/b + 3
5 – 3 = a ² /b ² - 2a√3/b
2 = a ² /b ² - 2a√3/b
2a√3/b = a ² /b ² - 2
2a√3/b = (a ² - 2 b ² )/ b ²
2a√3 = (a ² - 2 b ² )b / b ²
√3 = (a ² - 2 b ² )/2ab
⇒ √3 is a rational number which contradicts our assumption.
So, √5 + √3 is an irrational number.

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Euclid's Geometry

• Euclid Geometry
• Euclids division lemma
• Euclids division Algorithm
• Fundamental Theorem of Arithmetic
• Finding HCF LCM of positive integers
• Proving Irrationality of Numbers
• Decimal expansion of Rational numbers

From Euclid Geometry to Real numbers

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