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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 : , , π, e ,... etc. are irrational numbers.


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

We know that
1 < 2 < 4 ,
⇒ 12 < 2 < 22
Taking positive square roots we get 1 < < 2
Also 1.96 < 2 < 2.25  ⇒   (1.4)2 <2 < (1.5)2
taking positive square roots, we get 1.4 < < 1.5
Next 1.9881 < 2 < 2.0164 ⇒  (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
⇒ (1.4142135)2 < 2 < (1.4142136)2
⇒ 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 non–repeating and non–terminating decimal expansion of  will be given by:
= 1.4142135 …

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 = p1p2 . . . pn where p1 , p2 , . . . pn are primes, not necessarily distinct.
Therefore , a2 = .( p1p2 . . . pn ) ( p1p2 . . . pn ) = p12p22 . . . pn2 .

Now, we are given that p divides a2. Therefore, from the fundamental theorem of arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a2 are p1 , p2 , . . .,pn . So p is one of the p1 , p2 , … .pn . Now, since a = p1p2 … pn, p divides a.

Definition :-
A number is irrational if and only if its decimal representation is non–terminating and non–repeating.

There are infinitely many irrational numbers. [There exists a point on line ‘l’ 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 ‘a’. Now let us examine as to where this irrational-number ‘a’ stands in relation to rational numbers.

0.1 < a < 0.2 where 0.1 and 0.2 are rational numbers. So, ‘a’ lies on the line–segment whose end–points are 0.1 and 0.2 on the number line ‘l’.

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

This shows that ‘a’ lies on a line–segment whose end–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 ‘l’. Since every segment contains ‘a’, therefore the point p will represent the irrational number ‘a’.


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 ‘l’.


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 = p1p2 . . . pn where p1 , p2 , . . . pn are primes, not necessarily distinct.
Therefore , a2 = .( p1p2 . . . pn ) ( p1p2 . . . pn ) = p12p22 . . . pn2 .
Now, we are given that p divides a2. Therefore, from the fundamental theorem of arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realize that the only prime factors of a2 are p1 , p2 , . . .,pn . So p is one of the p1 , p2 , … .pn . Now, since a = p1p2 … pn, p divides a.

Definition :-
A number is irrational if and only if its decimal representation is non–terminating and non–repeating.
There are infinitely many irrational numbers. [There exists a point on line ‘l’ 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 non–terminating decimal representation. In case of rational numbers, the decimal representation is repeating (including repeating zeroes) and if the decimal representation is non–repeating, it is an irrational number. For every real number, there corresponds a unique point on the number line ‘l’ or we may say that every point on the line ‘l’ 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 one–to–one correspondence between the real numbers and points on the number line ‘l’, that is why the number line is called the ‘real number line’. 

 

Example

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

Solution

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: - PN2 = PM2 + MN2
                  = 12 + 12 = 2
         ⇒ PN =
In D SPN, PS2 = PN2 + NS2
                       = ()2+ 12
                       = 3
  ⇒ PS = ]

 
Example

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

Solution

Let us assume 2 + √2 be a rational number say r.
Then 2 + √2 = r
√2 = r - 2
But we know that √2 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 + √2 is not a rational number.

 
Example

Prove that 3√3 is not a rational number.

Solution

Let us assume 3√3 be a rational number say r.
Then 3√3 = r
√3 = (1/3)r
(1/3) r is a rational number because product of two rational number is a rational number.
=> √3 is a rational number but we know that √3 is not a rational number.
Therefore our assumption that 3√3 is a rational number is wrong.

Hence 3√3 is not a rational number

 
Example

Show that is not a rational number.

Solution

Let be a rational number, say where q ≠ 0.
Then   =
Since 13 = 1 , and 23 = 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 = 
6q2
q being an integer, 6q2 is an integer, and since q > 1 and q does not have a common factor with p and consequently with p3. So, is a fraction different from an integer. 
Thus 6q2.
This contradiction proves the result.

 
Example

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

Solution

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.

 
Example

Find the value of √5 correct to two places of decimal.

Solution

We know that 22 = 4 < 5 < 9 = 32
Taking positive square roots we get
2 < √5 < 3.
Next, (2.2)2 = 4.84 < 5 < 5.29 = (2.3)2
Taking positive square roots, we have
2.2 < √5 < 2.3
Again, (2.23)2 = 4.9729< 5 < 5.0176 = ( 2.24)2
Taking positive square roots, we obtain
2.23 < √5 < 2.24
Hence the required approximation is 2.24 as (2.24)2 is nearer to 5 than (2.23)2

 
Example
  Prove that √3 - √2 is irrational.
Solution

Let √3 - √2 be a rational number, = r (say )
Then √3 - √2 = r
On squaring both sides we have
    (√3 - √2)2 = r2
  3 - 2√6 + 2 = r2 
        5 - 2√6 = r2
           - 2√6 = r2 - 5
               √6 = -(r2 - 5)/2
Now -(r2 - 5)/2 is a rational number but √ 6 is an irrational number. 
Since a rational number cannot be equal to an irrational number. Our assumption that
√3 - √2 is rational is wrong. 

Ηence √3 - √2 is irrational. 

 
Example

Prove that √3 + √5 is an irrational number.

Solution

Let √3 + √5 be a rational number, = r (say )
Then √3 + √5 = r
On squaring both sides, 
     (√3 + √5)2 = r2
 3 + 2√15 + 5 = r2 
       8 + 2√15 = r2
             2√15 = r2 - 8
               √15 = (r2 - 8)/2
Now (r2 - 8)/2 is a rational number but √ 15 is an irrational number. 
Since a rational number cannot be equal to an irrational number. Our assumption that
√3 + √5 is rational is wrong.

Ηence √3 + √5 is an irrational number.

 
Example

Prove that
(a) 2 + is not a rational number and
(b)  is not a rational number. 

Solution

If possible, let 2 + = a , where a is rational.
Then, (2 + )2= a2

7 + 4= a2

= -------(i)
Now, a is rational ⇒ 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 ≠ 0.

Then,    ()3 = (p/q)3

           ⇒ 7 = p3/q3
         ⇒ 7q = p------(i)
         ⇒ p is a multiple of 7
         ⇒ p is multiple of 7.

     Let p = 7m, where m is an integer.
Then, p3 = 343 m3 ------(ii)
   ⇒ 7q = 343 m3 [from (i) and (ii)]
     ⇒ q = 49 m3
        ⇒ q is a multiple of 7.
        ⇒ 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.






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