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An irrational number is any real number that cannot be

expressed as a ratio a/b, where a and b are integers, with

b non-zero, and therefore not a rational number.

We have studied many properties of irrational numbers,

we have also located irrational numbers on the number

line and have seen that every irrational number

corresponds to a definite point on the number line. But

we have not prove that they are irrational. Here we shall

prove that if ‘p’ is a prime number then √p is irrational.

For this we take help of the Fundamental Theorem of

Arithmetic.

• Let ‘p’ be a prime number. If p divides a², then p divides a

where a is a positive integer.

• Proof:- Given a is a positive integer. So, let p1 , p2 ,p3, ..

pn be prime factor (not necessarily all distinct) of a.

• Then a=p1.p2.p3 ….pn

• Therefore, a2 =(p1 p1.p2.p3 ….pn) (p1 p1.p2.p3….pn)

• =p12.p2

2.p32 ….pn

2

• It is given that p divides a².

• So, p is a prime factor of a², from the fundamental

theorem of arithmetic.

• But the only prime factors of a² are p1 , p2 ,p3, .. Pn.

Therefore p is one factor of p1 , p2 ,p3, .. Pn. Hence p

divides a.

• In this method we assume a contradictory statement of what is

to be prove and come to a conclusion that contradicts the

hypothesis. And so the theorem is proved. For example:

• By the method of contradiction prove that √2 is irrational.

• Solution: We assume that √2 is rational. So, in its simplest

form, let √2= p/q, where p and q are integers having no

common factor other than 1 and q≠0.

• Then p=q √2 → p² = 2q²

• From (1) p² is divisible by 2, so p is divisible by 2.

• Let p= 2m→ p²=(2m)² = 2q²

• 4m²= 2q² →2m² = q²

• Hence, q² is divisible by 2 and consequently q is divisible

by 2. So , p is divisible by 2 as well as q is divisible by 2.

• Thus, 2 is a common factor of p and q which contradicts

the assumption, that p and q have no common factor other

than 1.

• This contradiction is due to the incorrect assumption that

√2 is rational. Hence √2 is irrational.

• Show that 5-√3 is irrational.

• Solution: Assume that 5-√3 is rational.

• That is, we can find co prime a and b (b≠0) such that 5-

√3 =a/b.

• Therefore, 5-a = √3

b

• Rearranging this equation, we get √3= 5-a = 5b-a

b b

• Since a and b are integers we get 5-a is rational and so

• b

• √3 is rational. But this contradict the fact that √3 is

rational.

• This contradiction has arisen due to incorrect assumption

that 5- √3 is irrational.

• So 5-√3 is irrational.

• So we conclude that:-

• The sum or difference of a rational and irrational number

is irrational.