Proving Irrationality of Numbers

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.

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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