# Revisiting Irrational Numbers

Numbers which are not rational, are called irrational numbers. Those numbers which have non-terminating and non-repeating decimal representations are called irrational numbers.

**Example :** , , Ï€, e ,... etc. are irrational numbers.

In earlier classes, we have read that is an irrational number.

We know that

1 < 2 < 4 ,

â‡’ 1^{2} < 2 < 2^{2}

Taking positive square roots we get 1 < < 2

Also 1.96 < 2 < 2.25 â‡’ (1.4)^{2} <2 < (1.5)^{2}

taking positive square roots, we get 1.4 < < 1.5

Next 1.9881 < 2 < 2.0164 â‡’ (1.41)^{2} < 2 < (1.42)^{2}

taking positive square roots, we get 1.41 < < 1.42

If we continue this process, the next step will give the following inequalities:

1.414 < < 1.415

Proceeding in this manner, we find that every new step gives a closer decimal approximation of than the previous step. The eighth step will give the following inequalities:

1.99999982358225 < 2 < 2.00000010642496

â‡’ (1.4142135)^{2} < 2 < (1.4142136)^{2}

â‡’ 1.4142135 < < 1.4142136

We have already learnt is not a rational number. So the above process will never terminate and nor will it be repeating. Hence, the nonâ€“repeating and nonâ€“terminating decimal expansion of will be given by:

= 1.4142135 â€¦

By a similar method, we can show that decimal representation of, etc. will not terminate and nor will they be repeating.

Theorem 1.3 : Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.

**Proof:**

Let the prime factorization of a be as follows :

a = p_{1}p_{2} . . . p_{n} where p_{1} , p_{2} , . . . p_{n} are primes, not necessarily distinct.

Therefore , a^{2} = .( p_{1}p_{2} . . . p_{n} ) ( p_{1}p_{2} . . . p_{n} ) = p_{1}^{2}p_{2}^{2} . . . p_{n}^{2} .

Now, we are given that p divides a^{2}. Therefore, from the fundamental theorem of arithmetic, it follows that p is one of the prime factors of a^{2}. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a^{2} are p_{1} , p_{2} , . . .,p_{n} . So p is one of the p_{1} , p_{2} , â€¦ .p_{n} . Now, since a = p_{1}p_{2} â€¦ p_{n}, p divides a.

**Definition** **:-**

A number is irrational if and only if its decimal representation is nonâ€“terminating and nonâ€“repeating.

There are infinitely many irrational numbers. [There exists a point on line â€˜lâ€™ corresponding to every irrational number.]

# Representation of Irrational Numbers on Number Line

Let there be a irrational number 0.1010010001000001 and let it is denoted by â€˜aâ€™. Now let us examine as to where this irrational-number â€˜aâ€™ stands in relation to rational numbers.0.1 < a < 0.2 where 0.1 and 0.2 are rational numbers. So, â€˜aâ€™ lies on the lineâ€“segment whose endâ€“points are 0.1 and 0.2 on the number line â€˜lâ€™.

Again, 0.101 and 0.102 are rational numbers such that 0.101 < a < 0.102

This shows that â€˜aâ€™ lies on a lineâ€“segment whose endâ€“points are 0.101 and 0.102.

It is clear that second segment is completely contained in the first segment and the third segment is completely contained in the second segment. Thus we see that there is a succession of small segments such that each segment contains *a* and is contained in the previous segment. The lengths of these segments are continuously decreasing. Hence this succession of segments will ultimately shrink to a point *p* (say) on the line â€˜lâ€™. Since every segment contains â€˜*a*â€™, therefore the point *p* will represent the irrational number â€˜*a*â€™.

**From the above discussion, it can be concluded that:**

(i) An irrational number can be approximated, as closely as we like, by a rational number.

(ii) To an irrational number, there corresponds an unique point on the number line â€˜lâ€™.

Theorem 1.3 : Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.

**Proof:**

Let the prime factorization of a be as follows :

a = p_{1}p_{2} . . . p_{n} where p_{1} , p_{2} , . . . p_{n} are primes, not necessarily distinct.

Therefore , a^{2} = .( p_{1}p_{2} . . . p_{n} ) ( p_{1}p_{2} . . . p_{n} ) = p_{1}^{2}p_{2}^{2} . . . p_{n}^{2} .

Now, we are given that p divides a^{2}. Therefore, from the fundamental theorem of arithmetic, it follows that p is one of the prime factors of a^{2}. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a^{2} are p_{1} , p_{2} , . . .,p_{n} . So p is one of the p_{1} , p_{2} , â€¦ .p_{n} . Now, since a = p_{1}p_{2} â€¦ p_{n}, p divides a.

**Definition** **:-**

A number is irrational if and only if its decimal representation is nonâ€“terminating and nonâ€“repeating.

There are infinitely many irrational numbers. [There exists a point on line â€˜lâ€™ corresponding to every irrational number.]

# Real Numbers

Rational numbers and irrational numbers taken together form the set of real numbers. The set of real numbers is denoted by*R*. Thus every real number is either a rational number or an irrational number. In either case, it has a nonâ€“terminating decimal representation. In case of rational numbers, the decimal representation is repeating (including repeating zeroes) and if the decimal representation is nonâ€“repeating, it is an irrational number. For every real number, there corresponds a unique point on the number line â€˜lâ€™ or we may say that every point on the line â€˜lâ€™ corresponds to a real number (rational or irrational).

**From the above discussion we may conclude that:**

To every real number there corresponds a unique point on the number line and conversely, to every point on the number line there corresponds a real number. Thus we see that there is oneâ€“toâ€“one correspondence between the real numbers and points on the number line â€˜lâ€™, that is why the number line is called the â€˜

*real number line*â€™.

Determine the point on the number line which represents the irrational number .

Mark the points 0, 1, 2 on the number line l. Let P denote the point 0 (zero) and M denote the point 1 (one). Draw a straight line MN perpendicular to at the point M and let N be a point on this perpendicular such that PM = MN = 1 unit in length.

Join PN, PN = units. Now draw NS perpendicular to PN such that NS = 1 unit of length. Now taking P as centre and radius equal to PS draw an arc intersecting the line l * * in R.

Then PR = and point R represents the irrational number because PR = PS = .

[Note: - PN

^{2}= PM

^{2}+ MN

^{2}

= 1

^{2}+ 1

^{2}= 2

⇒ PN =

In D SPN, PS

^{2}= PN

^{2}+ NS

^{2}

= ()

^{2}+ 1

^{2}

= 3

⇒ PS = ]

You have seen that âˆš2 is not a rational number. Show that 2 + âˆš2 is not a rational number.

Let us assume 2 + âˆš2 be a rational number say r.

Then 2 + âˆš2 = r

âˆš2 = r - 2

But we know that âˆš2 is an irrational number.

Therefore, r - 2 is also an irrational number.

=>r is an irrational number.

Hence our assumption r is a rational number is wrong.

Hence 2 + âˆš2 is not a rational number.

Prove that 3âˆš3 is not a rational number.

Let us assume 3âˆš3 be a rational number say r.

Then 3âˆš3 = r

âˆš3 = (1/3)r

(1/3) r is a rational number because product of two rational number is a rational number.

=> âˆš3 is a rational number but we know that âˆš3 is not a rational number.

Therefore our assumption that 3âˆš3 is a rational number is wrong.

Hence 3âˆš3 is not a rational number

Show that is not a rational number.

Let be a rational number, say where q â‰ 0.

Then ** = **

Since 1^{3} = 1 , and 2^{3} = 8, it follows that 1 < < 2

Then q > 1 because if q = 1 then will be an integer, and there is no integer between 1 and 2.

Now, 6 =

6 =

6q^{2} =

q being an integer, 6q^{2} is an integer, and since q > 1 and q does not have a common factor with p and consequently with p^{3}. So, is a fraction different from an integer.

Thus 6q^{2}â‰ .

This contradiction proves the result.

Give two examples to show that the product of two irrational numbers may be a rational number.

Take a = (2 + ) and b =(2 - ); a and b are irrational numbers,

but their product = 4 - 3 =1,is a rational number.

Take c = and d = -; c and d are irrational numbers, but their product = -3,

is a rational number.

Find the value of âˆš5 correct to two places of decimal.

We know that 2^{2} = 4 < 5 < 9 = 3^{2}

Taking positive square roots we get

2 < âˆš5 < 3.

Next, (2.2)^{2} = 4.84 < 5 < 5.29 = (2.3)^{2}

Taking positive square roots, we have

2.2 < âˆš5 < 2.3

Again, (2.23)^{2} = 4.9729< 5 < 5.0176 = ( 2.24)^{2}

Taking positive square roots, we obtain

2.23 < âˆš5 < 2.24

Hence the required approximation is 2.24 as (2.24)^{2} is nearer to 5 than (2.23)^{2} .

Let âˆš3 - âˆš2 be a rational number, = r (say )

Then âˆš3 - âˆš2 = r

On squaring both sides we have

(âˆš3 - âˆš2)^{2} = r^{2}

3 - 2âˆš6 + 2 = r^{2}

5 - 2âˆš6 = r^{2}

- 2âˆš6 = r^{2} - 5

âˆš6 = -(r^{2} - 5)/2

Now -(r^{2} - 5)/2 is a rational number but âˆš 6 is an irrational number.

Since a rational number cannot be equal to an irrational number. Our assumption that

âˆš3 - âˆš2 is rational is wrong.

Î—ence âˆš3 - âˆš2 is irrational.

Prove that âˆš3 + âˆš5 is an irrational number.

Let âˆš3 + âˆš5 be a rational number, = r (say )

Then âˆš3 + âˆš5 = r

On squaring both sides,

(âˆš3 + âˆš5)^{2} = r^{2}

3 + 2âˆš15 + 5 = r^{2}

8 + 2âˆš15 = r^{2}

2âˆš15 = r^{2} - 8

âˆš15 = (r^{2} - 8)/2

Now (r^{2} - 8)/2 is a rational number but âˆš 15 is an irrational number.

Since a rational number cannot be equal to an irrational number. Our assumption that

âˆš3 + âˆš5 is rational is wrong.

Î—ence âˆš3 + âˆš5 is an irrational number.

Prove that

(a) 2 + is not a rational number and

(b) is not a rational number.

If possible, let 2 + = a , where a is rational.

Then, (2 + )^{2}= a^{2}

7 + 4= a^{2}

= -------(i)

Now, a is rational â‡’ is rational.

is rational [from (i)]

This is a contradiction (as is irrational).

Hence, 2 + is not a rational number.

(b) If possible, let **=** p/q , where p and q are integers, having no common factors and q â‰ 0.

Then,** **()^{3} = (p/q)^{3}

â‡’ 7 = p^{3}/q^{3}

â‡’ 7q = p------(i)

â‡’ p is a multiple of 7

â‡’ p is multiple of 7.

Let p = 7m, where m is an integer.

Then, p^{3} = 343 m^{3} ------(ii)

â‡’ 7q = 343 m^{3} [from (i) and (ii)]

â‡’ q = 49 m^{3}

â‡’ q is a multiple of 7.

â‡’ q is a multiple of 7.

Thus, p and q are both multiples of 7, or 7 is a factor of p and q.

This contradicts our assumption that p and q have no common factors.

Hence is not a rational number.