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.
Question 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.
Question 2 :
Prove that √3 is an irrational number.
Solution :
Let √3 be a rational number.
Then it may be in the form a/b
√3 = a/b
Taking squares on both sides, we get
3 = a^{2}/b^{2}
3b^{2 }= a^{2}
a^{2} divides 3 (That is 3/a^{2})
Then a also divides 3.
Let a = 3c
3b^{2 }= a^{2}
By applying the value here, we get
3b^{2 }= (3c)^{2}
3b^{2 }= 9c^{2}
b^{2 }= 3c^{2}
b^{2} divides 3 (That is 3/b^{2})
Then b also divides 3.
From this, we come to know that a and b have common divisor other than 1. It means our assumption is wrong. Hence √3 is irrational.
Question 3 :
Prove that 3√2 is a irrational.
Solution :
Let us assume 3√2 as rational.
3√2 = a/b
√2 = a/3b
Since √2 is irrational
Since 3, a and b are integers a/3b be a irrational number. So it contradicts.
Hence 3√2 is irrational number.
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