Revisiting Irrational Numbers

CBSE Class 10 Mathematics-Real Numbers- Revisiting Irrational Numbers Notes with Examples. Irrational numbers can not be expressed in p/q form.

Revisiting Irrational Numbers

 

 

 

Revisiting Irrational Numbers

Irrational numbers are those numbers which cannot be written in the form {\color{Blue} \frac{p}{q}} , where p and q are integers and q ≠ 0. E.g.{\color{Blue} \sqrt{2}},{\color{Blue} \sqrt{3}}{\color{Blue} \sqrt{15}}

The square roots of all the numbers do not give an irrational number.

For example, {\color{Blue} \sqrt{2}}  is an irrational number but {\color{Blue} \sqrt{4}} = 2, which is rational.

Therefore square roots of all prime numbers are irrational.

If p is a prime number then {\color{Blue} \sqrt{p}} 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)(p1p2p... ... ... . . pn)

                                 a2 = (p12p22p32 ... ... ... . . pn2 )

It is given that p divides a. 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 aare 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 {\color{Red} \sqrt{2}}  is an irrational number.

Let us assume that {\color{Blue} \sqrt{2}} is rational. Since it is a rational number it can be expressed in the form  {\color{Blue} \frac{a}{b}}  , where a and b are integers and b ≠ 0.

Now, a and b have no common factor other than 1.
{\color{Blue} \sqrt{2}} = {\color{Blue} \frac{a}{b}} , where a and b are coprime

On squaring both sides, we get
                                  2 =  {\color{Blue} \frac{a^{2}}{b^{2}}}

                               2b2 = a2

Therefore, 2 divides a2.

We know that when 2 divides athen 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 b, 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 {\color{Blue} \sqrt{2}}  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 {\color{Blue} \frac{a}{b}} where a and b are integers and b ≠ 0. Now, a and b have no common factor other than 1.

3{\color{Blue} \sqrt{2}} {\color{Blue} \frac{a}{b}} , where a and b are coprime integers

{\color{Blue} \sqrt{2}} = {\color{Blue} \frac{a}{3b}}

Since, 3, a and b are integers, {\color{Blue} \frac{a}{3b}} is rational.

Now, a rational number cannot be equal to an irrational number, that is{\color{Blue} \sqrt{2}}

So, we conclude that 3{\color{Blue} \sqrt{2}} is irrational.

Example: Show that 5 − {\color{Red} \sqrt{3}} is irrational.

Let us assume that 5 −{\color{Blue} \sqrt{3}} is rational. As it is a rational number, it can be

expressed in the form {\color{Blue} \frac{a}{b}} where a and b are integers and b ≠ 0. Now, a and b have no common factor other than 1.

5 − {\color{Blue} \sqrt{3}} =ab, where a and b are coprime

5 −{\color{Blue} \frac{a}{b}}{\color{Blue} \sqrt{3}}

{\color{Blue} \frac{5b - a}b{}}{\color{Blue} \sqrt{3}}

As, 5, a, and b are integers, {\color{Blue} \frac{5b - a}b{}}  is rational.

Now, {\color{Blue} \sqrt{3}} is an irrational number and cannot be equal to a rational number.

So, we conclude that 5 − {\color{Blue} \sqrt{3}} is irrational.

Example: Prove that 2√3 + √5 is an irrational number.

Let us assume that 2{\color{Blue} \sqrt{3}} + √5 is rational. As it is a rational number, it can be expressed in the form {\color{Blue} \frac{a}{b}} where a and b are integers and b ≠ 0. Now, a  and b have no common  factor other than 1.

2{\color{Blue} \sqrt{3}} + {\color{Blue} \sqrt{5}} = {\color{Blue} \frac{a}{b}} , where a and b are coprime

2{\color{Blue} \sqrt{3}}{\color{Blue} \frac{a}{b}} − {\color{Blue} \sqrt{5}}

On squaring both sides, we get
12 = ( 
{\color{Blue} \frac{a}{b}}− √5 )2

12 =a2/b2 − 2{\color{Blue} \sqrt{5}} {\color{Blue} \frac{a}{b}}+ 5

{\color{Blue} \frac{a^{2}}{b^{2}}} − 2{\color{Blue} \sqrt{5}} {\color{Blue} \frac{a}{b}}= 7

{\color{Blue} \frac{a^{2}}{b^{2}}} − 7 = 2{\color{Blue} \sqrt{5}} {\color{Blue} \frac{a}{b}} 

{\color{Blue} \frac{a^{2} - 7b^{2}}{2b^{2}}} = {\color{Blue} \sqrt{5}} {\color{Blue} \frac{a}{b}}

{\color{Blue} \sqrt{5}} ={\color{Blue} \frac{a^{2} - 7b^{2}}{2ab}}
Since, 2, 7, a and b are integers {\color{Blue} \frac{a^{2} - 7b^{2}}{2ab}} is rational

A rational number cannot be equal to an irrational number.

So, our assumption is wrong. Hence, 2{\color{Blue} \sqrt{3}} + {\color{Blue} \sqrt{5}} is irrational.

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