Revisiting Irrational Numbers CBSE Class 10 Mathematics Notes with Examples

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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 ... ... ... . . p_n where p₁p₂p₃ ... ... ... . . p_n are all the prime numbers of a.

a² = (p1p2p3 ... ... ... . . pn)

a² = (p₁p₂p₃ ... ... ... . . pn)(p₁p₂p₃... ... ... . . p_n)

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 a²are p₁p₂p₃ ... ... ... . . p_n. 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 a²then 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 ${\color{Blue} \sqrt{2}}$ 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