- Books Name
- Mathematics Book for CBSE Class 10
- Publication
- Carrier Point
- Course
- CBSE Class 10
- Subject
- Mathmatics
Revisiting Irrational Numbers
In Class IX, you were introduced to irrational numbers and many of their properties.
You studied about their existence and how the rationals and the irrationals together
made up the real numbers. You even studied how to locate irrationals on the number
line. However, we did not prove that they were irrationals. In this section, we will
prove that 2 , 3 , 5 and, in general, p is irrational, where p is a prime. One of
the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
Recall, a number ‘s’ is called irrational if it cannot be written in the form ,
p
q
where p and q are integers and q ¹ 0. Some examples of irrational numbers, with
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12 MATHEMATICS
which you are already familiar, are :
2 2, 3, 15 , , , 0.10110111011110 . . .
3
p − , etc.
Before we prove that 2 is irrational, we need the following theorem, whose
proof is based on the Fundamental Theorem of Arithmetic.
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 factorisation of a be as follows :
a = p1p2 . . . pn, where p1,p2, . . ., pn are primes, not necessarily distinct.
Therefore, a2 = (p1p2 . . . pn)( p1 p 2 . . . pn) = p2
1p2
2 . . . p2
n.
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 realise that the only
prime factors of a2 are p
1, p
2, . . ., p
n. So p is one of p
1, p
2, . . ., p
n.
Now, since a = p1 p2 . . . pn, p divides a.
We are now ready to give a proof that 2 is irrational.
The proof is based on a technique called ‘proof by contradiction’. (This technique is
discussed in some detail in Appendix 1).
Theorem 1.4 : 2 is irrational.
Proof : Let us assume, to the contrary, that 2 is rational.
So, we can find integers r and s (¹ 0) such that 2 =
r
s
.
Suppose r and s have a common factor other than 1. Then, we divide by the common
factor to get 2 ,
a
b
= where a and b are coprime.
So, b 2 = a.
Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2.
Now, by Theorem 1.3, it follows that 2 divides a.
So, we can write a = 2c for some integer c.
* Not from the examination point of view.
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REAL NUMBERS 13
Substituting for a, we get 2b2 = 4c2, that is, b2 = 2c2.
This means that 2 divides b2, and so 2 divides b (again using Theorem 1.3 with p = 2).
Therefore, a and b have at least 2 as a common factor.
But this contradicts the fact that a and b have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that 2 is rational.
So, we conclude that 2 is irrational.
Example 9 : Prove that 3 is irrational.
Solution : Let us assume, to the contrary, that 3 is rational.
That is, we can find integers a and b (¹ 0) such that 3 =
a
b
×
Suppose a and b have a common factor other than 1, then we can divide by the
common factor, and assume that a and b are coprime.
So, b 3 = a×
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible
by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3
with p = 3).
Therefore, a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that 3 is irrational.
In Class IX, we mentioned that :
l the sum or difference of a rational and an irrational number is irrational and
l the product and quotient of a non-zero rational and irrational number is
irrational.
We prove some particular cases here.
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14 MATHEMATICS
Example 10 : Show that 5 – 3 is irrational.
Solution : Let us assume, to the contrary, that 5 – 3 is rational.
That is, we can find coprime a and b (b ¹ 0) such that 5 3
a
b
− = ×
Therefore, 5 3
a
b
− = ×
Rearranging this equation, we get
5
3 5 –
a b a
b b
−
= = ×
Since a and b are integers, we get 5 –
a
b
is rational, and so 3 is rational.
But this contradicts the fact that 3 is irrational.
This contradiction has arisen because of our incorrect assumption that 5 – 3 is
rational.
So, we conclude that 5 − 3 is irrational.
Example 11 : Show that 3 2 is irrational.
Solution : Let us assume, to the contrary, that 3 2 is rational.
That is, we can find coprime a and b (b ¹ 0) such that 3 2
a
b
= ×
Rearranging, we get 2
3
a
b
= ×
Since 3, a and b are integers,
3
a
b
is rational, and so 2 is rational.
But this contradicts the fact that 2 is irrational.
So, we conclude that 3 2 is irrational.
- Books Name
- Rakhiedu Mathematics Book
- Publication
- Param Publication
- Course
- CBSE Class 10
- Subject
- Mathmatics
1.3 Irrational Numbers :
Some Properties of irrational numbers :
(a) The –ve of an irrational number is an irrational number.
(b) The sum of a rational and an irrational number is an irrational number.
(c) The product of a non–zero rational number with an irrational number is always an irrational number.
Irrational Numbers
We know that a number which cannot be written in the form of , where p and q are integers and q ≠ 0, is known as an irrational number.
For example: all numbers of the form , where p is a prime number such as etc., are irrational numbers.
How can we prove that these are irrational numbers?
We can prove this by making use of a theorem which can be stated as follows.
“If p divides a2, then p divides a (where p is a prime number and a is a positive integer)”.
So go through the given video to understand the application of the above stated property.
Similarly, we can prove that square roots of other prime numbers like , etc. are irrational numbers.
Besides these irrational numbers, there are some other irrational numbers like etc.
We can also prove why these numbers are irrational. Before this, let us first see what happens to irrational numbers, when we apply certain mathematical operations on them.
We will now prove that is irrational.
We know that is irrational (as proved before).
Now, the multiplication of a rational and an irrational number gives an irrational number. Therefore, is an irrational number.
Let us now try to understand the concept further through some more examples.
Example 1: Prove that is irrational. Solution:
Let us assume that is not irrational, i.e. is a rational number.
Then we can write , where a and b are integers andb ≠ 0.
Let a and b have a common factor other than 1.
After dividing by the common factor, we obtain
, where c and d are co-prime numbers.
As c, d and 2 are integers, and are rational numbers.
Thus, is rational.
is rational as the difference of two rational numbers is again a rational number.
This is a contradiction as is irrational.
Therefore, our assumption that is rational is wrong.
Hence, is irrational.
Example 2: Prove that is irrational.
Solution:
Let us assume is rational. Then, we can write
,
where a and b are co-prime and b ≠ 0.
⇒
Now, as a and b are integers, is rational or is a rational number.
This means that is rational.
This is a contradiction as is irrational.
Therefore, our assumption that is rational is wrong.
Hence, is an irrational number.
Decimal Expansions of Rational Numbers
The Need for Converting Rational Numbers into Decimals
A carpenter wishes to make a point on the edge of a wooden plank at 95 mm from any end. He has a centimeter tape, but how can he use that to mark the required point?
Simple! He should convert 95 mm into its corresponding centimeter value, i.e., 9.5 cm and then measure and mark the required length on the wooden plank.
This is just one of the many situations in life when we face the need to convert numbers into decimals. In this lesson, we will learn to convert rational numbers into decimals, observe the types of decimal numbers, and solve a few examples based on this concept.
Know More
Two rational numbers and are equal if and only if ad = bc.
Take, for example, the rational numbers and . Let us see if they are equal or not. Here, a = 2, b = 4, c = 3 and d = 6
Now, we have:
ad = 2 × 6 = 12
bc = 4 × 3 = 12
Since ad = bc, we obtain = .
We know that the form represents the division of integer p by the integer q. By solving
this division, we can find the decimal equivalent of the rational number . Now, let us convert the numbers , and into decimals using the long division method.
While the remainder is zero in the division of 5 by 8, it is not so in case of the other two divisions. Thus, we can get two different cases in the decimal expansions of rational numbers.
Observing the Decimal Expansions of Rational Numbers
We can get the following two cases in the decimal expansions of rational numbers.
Case I: When the remainder is zero
In this case, the remainder becomes zero and the quotient or decimal expansion terminates after a finite number of digits after the decimal point. For example, in the decimal
expansion of , we get the remainder as zero and the quotient as 0.625.
Case II: When the remainder is never zero
In this case, the remainder never becomes zero and the corresponding decimal expansion
is non-terminating. For example, in the decimal expansions of and , we see that the remainder never becomes zero and their corresponding quotients are non-terminating decimals.
When we divide 4 by 3 and 2 by 7, we get 1.3333… and 0.285714285714… as the respective quotients. In these decimal numbers, the digit ‘3’ and the group of digits
‘285714’ get repeated. Therefore, we can write and
Here, the symbol indicates the digit or group of digits
that gets repeated.
Solved Examples
Example 1: Write the decimal expansion of and find if it is terminating or non- terminating and repeating.
Solution:
Here is the long division method to find the decimal expansion of .
Hence, the decimal expansion of is 49.48. Since the remainder is obtained as zero, the decimal numberis terminating.
Example 2: Write the decimal expansion of and find if it is terminating or non- terminating and repeating.
Solution:
Here is the long division method to find the decimal expansion of .
Hence, the decimal expansion of is 87.33 Since the remainder 9 is obtained again
and again, the decimal numberis non-terminating and repeating. The decimal number can also be written as .
Medium
Example 1: Find the decimal expansion of each of the following rational numbers and write the nature of the same.
1.
2.
3.
4.
Solution:
We have = 0.64356435... =
The group of digits ‘6435’ repeats after the decimal point. Hence, the decimal expansion of the given rational number is non-terminating and repeating.
We have = 2.3075
Hence, the given rational number has a terminating decimal expansion.
We have = 0.3737... =
The pair of digits ‘37’ repeats after the decimal point. Hence, the decimal expansion of the given rational number is non-terminating and repeating.
We have = 0.67
Hence, the given rational number has a terminating decimal expansion.
Terminating and Non-terminating Repeating Decimal Expansions of Rational Numbers
We can find the decimal expansion of rational numbers using long division method.
However, it is possible to check whether the decimal expansion is terminating or non- terminating without actually carrying out long division also.
Let us start by taking a few rational numbers in the decimal form.
(a)
(b)
0.275
On prime factorizing the numerator and the denominator, we obtain
Can you see a pattern in the two examples?
We notice that the given examples are rational numbers with terminating decimal expansions. When they are written in the form, where p and q are co-prime
(the HCF of p and q is 1), the denominator, when written in the form of prime factors, has 2 or 5 or both.
The above observation brings us to the given theorem.
If x is a rational number with terminating decimal expansion, then it can be expressedin the form, where p and q are co-prime (the HCF of p and q is 1) and the prime factorisation of q is of the form 2n5m, where n and m are non-negative integers.
Contrary to this, if the prime factorisation of q is not of the form 2n5m, where n and m are non-negative integers, then the decimal expansion is a
non-terminating one.
Let us see a few examples that will help verify this theorem.
(a)
(b)
(c)
(d)
Note that in examples (b) and (d), each of the denominators is composed only of the prime factors 2 and 5, because of which, the decimal expansion is terminating. However, in examples (a) and (c), each of the denominators has at least one prime factor other than 2 and 5 in their prime factorisation, because of which, the decimal expansion is non- terminating and repetitive.
To summarize the above results, we can say that:
Let us solve a few examples to understand this concept better.
Example 1: Without carrying out the actual division, find if the following rational numbers have a terminating or a non-terminating decimal expansion.
(a)
(b)
Solution:
(a)
As the denominator can be written in the form 2n5m, where n = 6 and m = 2 are non-negative integers, the given rational number has a terminating decimal expansion.
(b)
As denominator cannot be written in the form 2n5m, where n and m are non-negative integers, the given rational number has a non-terminating decimal expansion.
Example 2: Without carrying out the actual division, find if the expression has a terminating or a non-terminating decimal expansion.
Solution:=
As the denominator can be written in the form 2n5m, where n = 7and m = 0 are non-negative integers, the given rational number has a terminating decimal expansion.
Hence, 5.5859375 is the decimal expansion of the given rational number.
- Books Name
- Sample paper Term II Maths
- Publication
- SonikaAnandAcademy
- Course
- CBSE Class 10
- Subject
- Mathmatics
There are two types of Real numbers
Rational
Irrrational
Rational numbers are of form p/q and terminating. Eg 5/7
Irrational numbers are repeated eg √2+5