A real number that is not rational is called an irrational number.
Theorem to Remember :
Let p be a prime number and a be a positive integer. If p divides a^{2}, then p divides a.
Example 1 :
Prove that √2 is an irrational number.
Solution :
Let √2 be a rational number.
Then it may be in the form a/b
√2 = a/b
Taking squares on both sides, we get
2 = a^{2}/b^{2}
2b^{2 }= a^{2}
a^{2} divides 2 (That is 2/a^{2})
Then a also divides 2.
Let a = 2c
2b^{2 }= a^{2}
By applying the value here, we get
2b^{2 }= (2c)^{2}
2b^{2 }= 4c^{2}
b^{2 }= 2c^{2}
b^{2} divides 2 (That is 2/b^{2})
Then b also divides 2.
From this, we come to know that a and b have common divisor other than 1. It means our assumption is wrong. Hence √2 is irrational.
Example 2 :
5 - √3 is irrational.
Solution :
Let 5 - √3 be a rational number.
Then it may be in the form a/b
5 - √3 = a/b
Taking squares on both sides, we get
5 - (a/b) = √3
(5b - a)/b = √3
a, b and 5 are rational numbers. Then the simplified value of (5b - a)/b must be rational. But it is clear that √3 is irrational.
So, it contradicts our assumption. Hence 5 - √3 is irrational.
Example 3 :
3 + 2√5 is irrational.
Solution :
Let 3 + 2√5 be a rational number.
Then it may be in the form a/b
3 + 2√5 = a/b
Taking squares on both sides, we get
3 - (a/b) = 2√5
(3b - a)/b = 2√5
(3b - a)/2b = √5
a, b, 3 and 2 are rational numbers. Then the simplified value of (3b - a)/2b must be rational. But it is clear that √5 is irrational.
So, it contradicts our assumption. Hence 3 + 2√5 is irrational.
Example 4 :
√2 + √5 is irrational.
Solution :
Let √2 + √5 be a rational number.
Then it may be in the form a/b
√2 + √5 = a/b
Taking squares on both sides, we get
2 + 5 + 2√10 = a^{2}/b^{2}
7 + 2√10 = a^{2}/b^{2}
2√10 = (a^{2}/b^{2}) - 7
2√10 = (a^{2}- 7b^{2})/b^{2}
√10 = (a^{2}- 7b^{2})/2b^{2}
a, b, 7 and 2 are rational numbers. Then the simplified value of (a^{2}- 7b^{2})/2b^{2} must be rational. But it is clear that √10 is irrational.
So, it contradicts our assumption. Hence √2 + √5 is irrational.
Example 5 :
3√2
Solution :
Let 3√2 be a rational number.
Then it may be in the form a/b
3√2 = a/b
√2 = a/3b
a, b and 3 are rational numbers. Then the simplified value of a/3b must be rational. But it is clear that √2 is irrational.
So, it contradicts our assumption. Hence 3√2 is irrational.
From the above examples, we come to know that
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