Revisiting Irrational Numbers
Revisiting Irrational Numbers
Irrational numbers are those numbers which cannot be written in the form , where p and q are integers and q ≠ 0. E.g.
,
,
The square roots of all the numbers do not give an irrational number.
For example, is an irrational number but
= 2, which is rational.
Therefore square roots of all prime numbers are irrational.
If p is a prime number then is an irrational number.
Theorem 1:
If a prime number p divides a2, then p divides a, where a is a positive integer.
Proof:
Every positive integer can be expressed as the product of primes.
Let a = p1p2p3 ... ... ... . . pn where p1p2p3 ... ... ... . . pn are all the prime numbers of a.
a2 = (p1p2p3 ... ... ... . . pn)
a2 = (p1p2p3 ... ... ... . . pn)(p1p2p3 ... ... ... . . pn)
a2 = (p12p22p32 ... ... ... . . pn2 )
It is given that p divides a2 . According to the Fundamental Theorem of Arithmetic, we can say that p is one of the prime factors of a2.
According to the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique.
Now the only prime factors of a2 are p1p2p3 ... ... ... . . pn. Therefore, p is one of p1p2p3 ... ... ... . . pn.
So p is also a factor of a
If p divides a2 then p also divides a.
Let us consider a positive integer 12.
Now, the factors of 12 are 2, 2 and 3.
On squaring 12 we get,
122 = 144
The factors of 144 are 2, 2, 2, 2, 3 and 3.
If one of the factors, let’s say 3 divides 144. Then this factor, 3 will also divide 144, as it is one of the factors of 144 also.
Example: Prove that is an irrational number.
Let us assume that is rational. Since it is a rational number it can be expressed in the form
, where a and b are integers and b ≠ 0.
Now, a and b have no common factor other than 1.
=
, where a and b are coprime
On squaring both sides, we get
2 =
2b2 = a2
Therefore, 2 divides a2.
We know that when 2 divides a2 then 2 divides a also.
We can write a = 2m, where m is an integer.
Putting a = 2m in 2b2 = a2
2b2 = (2m)2
2b2 = 4m2
b2 = 2m2
Again if 2 divides b2 , then 2 divides b also.
Therefore, 2 is a common factor of a and b.
But this contradicts the fact that a, b have no common factor other than 1.
So, we conclude that is irrational.
Example: Show that 3√2 is an irrational number.
Let us assume that 3√2 is rational.
Then 3√2 can be expressed in the form where a and b are integers and b ≠ 0. Now, a and b have no common factor other than 1.
3 =
, where a and b are coprime integers
=
Since, 3, a and b are integers, is rational.
Now, a rational number cannot be equal to an irrational number, that is
So, we conclude that 3 is irrational.
Example: Show that 5 − is irrational.
Let us assume that 5 − is rational. As it is a rational number, it can be
expressed in the form where a and b are integers and b ≠ 0. Now, a and b have no common factor other than 1.
5 − =ab, where a and b are coprime
5 −=
=
As, 5, a, and b are integers, is rational.
Now, is an irrational number and cannot be equal to a rational number.
So, we conclude that 5 − is irrational.
Example: Prove that 2√3 + √5 is an irrational number.
Let us assume that 2 + √5 is rational. As it is a rational number, it can be expressed in the form
where a and b are integers and b ≠ 0. Now, a and b have no common factor other than 1.
2 +
=
, where a and b are coprime
2=
−
On squaring both sides, we get
12 = ( − √5 )2
12 =a2/b2 − 2
+ 5
− 2
= 7
− 7 = 2
=
=
Since, 2, 7, a and b are integers is rational
A rational number cannot be equal to an irrational number.
So, our assumption is wrong. Hence, 2 +
is irrational.