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
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