In the previous section we saw how rational numbers are represented on a number line. In this section we will see irrational numbers.
We have seen examples of irrational numbers in an earlier chapter. We were able to represent √2 on the number line. We saw it in fig.16.6. For convenience it is shown again in fig.22.4(a) below.
In the same manner we can represent √3, √5 √7 etc., √3 is shown in fig.b
Now we come to the question of accuracy. We know that √2 = 1.41421356...
Can we mark a non-terminating non-recurring decimal like 1.41421356... accurately?
The answer is yes. The method is the same 'zooming in' that we saw in the fig.22.2 in the previous section. Let us see the details:
We know that 1.41421356... lies in between 1 and 2. In the fig.22.5 below, ‘zoom level 1’ shows the portion between 1 and 2, at an enlarged scale.
In this zoom level, we have enough space to clearly mark ten subdivisions between 1 and 2. So the fourth subdivision will be 1.4 and the fifth subdivision will be 1.5.
Then we zoom in on the region between this 1.4 and 1.5. This is shown as 'zoom level 2'. In this zoom level, we have enough space to clearly mark ten subdivisions between 1.4 and 1.5. So the first subdivision will be 1.41 and the second subdivision will be 1.42.
Then we zoom in on the region between this 1.41 and 1.42. This is shown as 'zoom level 3'. In this zoom level, we have enough space to clearly mark ten subdivisions between 1.41 and 1.42. So the fourth subdivision will be 1.414 and the fifth subdivision will be 1.415.
So, if we take the left red subdivision in zoom level 3, it will represent 1.414. It is fairly accurate. We know that 0.004 is a small quantity. But we are able to mark it by 'zooming in'. We can mark a quantity, even if it is non-recurring and non-terminating, however small it may be. All we need to do, is to zoom in to the required level.
Like √3, √7, π etc., also have decimal forms. Even though they are non-recurring and non-terminating decimals, they can be represented on a number line.
So we can conclude that:
Any irrational number can be represented on a number line.
Now we try to do the reverse: We are given an irrational number on the number line. Can we represent it as a fraction?
We have seen examples of irrational numbers in an earlier chapter. We were able to represent √2 on the number line. We saw it in fig.16.6. For convenience it is shown again in fig.22.4(a) below.
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| Fig.22.4 |
Now we come to the question of accuracy. We know that √2 = 1.41421356...
Can we mark a non-terminating non-recurring decimal like 1.41421356... accurately?
The answer is yes. The method is the same 'zooming in' that we saw in the fig.22.2 in the previous section. Let us see the details:
We know that 1.41421356... lies in between 1 and 2. In the fig.22.5 below, ‘zoom level 1’ shows the portion between 1 and 2, at an enlarged scale.
In this zoom level, we have enough space to clearly mark ten subdivisions between 1 and 2. So the fourth subdivision will be 1.4 and the fifth subdivision will be 1.5.
Then we zoom in on the region between this 1.4 and 1.5. This is shown as 'zoom level 2'. In this zoom level, we have enough space to clearly mark ten subdivisions between 1.4 and 1.5. So the first subdivision will be 1.41 and the second subdivision will be 1.42.
Then we zoom in on the region between this 1.41 and 1.42. This is shown as 'zoom level 3'. In this zoom level, we have enough space to clearly mark ten subdivisions between 1.41 and 1.42. So the fourth subdivision will be 1.414 and the fifth subdivision will be 1.415.
So, if we take the left red subdivision in zoom level 3, it will represent 1.414. It is fairly accurate. We know that 0.004 is a small quantity. But we are able to mark it by 'zooming in'. We can mark a quantity, even if it is non-recurring and non-terminating, however small it may be. All we need to do, is to zoom in to the required level.
Like √3, √7, π etc., also have decimal forms. Even though they are non-recurring and non-terminating decimals, they can be represented on a number line.
So we can conclude that:
Any irrational number can be represented on a number line.
The answer is:
No irrational numbers cannot be expressed in the form p⁄q. We have seen the proof for this, in the case of √2. (Details here)
We will now see some important differences between rational and irrational numbers:
I. Sum of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their sum is:
p⁄q + r⁄s = ps⁄qs + rq⁄sq = (ps+rq)⁄qs.
(b) Let us analyse the above result.
• p⁄q and r⁄s are rational numbers. So p, q, r and s are integers (... -4, -3, -2, -1, 0, 1, 2, 3, 4...)
• Also q and s not equal to zero
(c) So ps, rq and qs will be integers. (ps+rq) will also be an integer
(d) Thus, the result in (a) is an [ integer⁄integer]. That means, the result in (a) is a rational number
II The sum of two irrational numbers may be rational or irrational
Example 1: Sum of √3 and √2 is irrational
Proof:
5. If 'a' is a rational number, [a + 1⁄a] will also be a rational number.
6. Then 1⁄2[a + 1⁄a] will also be a rational number
7. But from (4), we have '1⁄2[a + 1⁄a]' is equal to √3 which is an irrational number.
8. So 'a' is irrational. That means (√3+ √2) is irrational
Example 2: Sum of (1+√2) and (1-√2) is rational
Proof:
• This can be proved just by adding: 1 + √2 + 1 - √2 = 2
• '2' is a rational number
III. Difference of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their difference is:
p⁄q - r⁄s = ps⁄qs - rq⁄sq = (ps-rq)⁄qs.
(b) From (I) above, we can write:
The result in (a) is an [ integer⁄integer]. That means, the result in (a) is a rational number
IV. The difference of two irrational numbers may be rational or irrational
This can be proved in the same way as in (II)
V. Product of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their product is:
(p⁄q) × (r⁄s) = pr⁄qs. This is a rational number
VI. The product of two irrational numbers may be rational or irrational
Example 1: Consider 6√5.
1. We have √5 = 2.236...
2. So 6√5 = 6 × 2.236... = 13.416...
3. 13.416... is an irrational number. So 6√5 is an irrational number
4. Similarly we can prove 2√5 is also an irrational number
5. Now take the products. 6√5 × 2√5 = 12 × 5 = 60
6. 60 is a rational number. So we got a rational number when two irrationals were multiplied
Example 2:
1. √3 × √5 = √15
2. √15 is an irrational number
3. So we got an irrational number when we multiplied two irrational numbers
VII. Quotient of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their quotient is:
(p⁄q) ÷ (r⁄s) = (p⁄q) × (s⁄r) = ps⁄qr. This is a rational number
VIII. The quotient of two irrational numbers may be rational or irrational
Example 1: (8√3)⁄(√3) = 8
'8' is a rational number
Example 2: (8√15)⁄(2√3) = (8√15)⁄(2√3) = (8×√5×√3)⁄(2√3) = 4√5
'4√5' is an irrational number
From I and II, we can conclude this:
• Sum of two rational numbers will always be a rational number.
• But sum of two irrational numbers may be rational or irrational
Conclusion 2:
From III and IV, we can conclude this:
• Difference of two rational numbers will always be a rational number.
• But difference of two irrational numbers may be rational or irrational
Conclusion 3:
From V and VI, we can conclude this:
• Product of two rational numbers will always be a rational number.
• But product of two irrational numbers may be rational or irrational
Conclusion 4:
From VII and VIII, we can conclude this:
• Quotient of two rational numbers will always be a rational number.
• But quotient of two irrational numbers may be rational or irrational
The above 4 conclusions are applicable when the rational numbers and irrational numbers are treated separately. We will now see what will be the result when they come together in calculations.
No irrational numbers cannot be expressed in the form p⁄q. We have seen the proof for this, in the case of √2. (Details here)
I. Sum of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their sum is:
p⁄q + r⁄s = ps⁄qs + rq⁄sq = (ps+rq)⁄qs.
(b) Let us analyse the above result.
• p⁄q and r⁄s are rational numbers. So p, q, r and s are integers (... -4, -3, -2, -1, 0, 1, 2, 3, 4...)
• Also q and s not equal to zero
(c) So ps, rq and qs will be integers. (ps+rq) will also be an integer
(d) Thus, the result in (a) is an [ integer⁄integer]. That means, the result in (a) is a rational number
II The sum of two irrational numbers may be rational or irrational
Example 1: Sum of √3 and √2 is irrational
Proof:
5. If 'a' is a rational number, [a + 1⁄a] will also be a rational number.
6. Then 1⁄2[a + 1⁄a] will also be a rational number
7. But from (4), we have '1⁄2[a + 1⁄a]' is equal to √3 which is an irrational number.
8. So 'a' is irrational. That means (√3
Example 2: Sum of (1+√2) and (1-√2) is rational
Proof:
• This can be proved just by adding: 1 + √2 + 1 - √2 = 2
• '2' is a rational number
III. Difference of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their difference is:
p⁄q - r⁄s = ps⁄qs - rq⁄sq = (ps-rq)⁄qs.
(b) From (I) above, we can write:
The result in (a) is an [ integer⁄integer]. That means, the result in (a) is a rational number
IV. The difference of two irrational numbers may be rational or irrational
This can be proved in the same way as in (II)
V. Product of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their product is:
(p⁄q) × (r⁄s) = pr⁄qs. This is a rational number
VI. The product of two irrational numbers may be rational or irrational
Example 1: Consider 6√5.
1. We have √5 = 2.236...
2. So 6√5 = 6 × 2.236... = 13.416...
3. 13.416... is an irrational number. So 6√5 is an irrational number
4. Similarly we can prove 2√5 is also an irrational number
5. Now take the products. 6√5 × 2√5 = 12 × 5 = 60
6. 60 is a rational number. So we got a rational number when two irrationals were multiplied
Example 2:
1. √3 × √5 = √15
2. √15 is an irrational number
3. So we got an irrational number when we multiplied two irrational numbers
VII. Quotient of two rational numbers will be a rational number.
Proof:
(a) Consider two rational numbers p⁄q and r⁄s. Their quotient is:
(p⁄q) ÷ (r⁄s) = (p⁄q) × (s⁄r) = ps⁄qr. This is a rational number
VIII. The quotient of two irrational numbers may be rational or irrational
Example 1: (8√3)⁄(√3) = 8
'8' is a rational number
Example 2: (8√15)⁄(2√3) = (8√15)⁄(2√3) = (8×√5×√3)⁄(2√3) = 4√5
'4√5' is an irrational number
From the above discussion, we can make the following conclusions:
Conclusion 1:From I and II, we can conclude this:
• Sum of two rational numbers will always be a rational number.
• But sum of two irrational numbers may be rational or irrational
Conclusion 2:
From III and IV, we can conclude this:
• Difference of two rational numbers will always be a rational number.
• But difference of two irrational numbers may be rational or irrational
Conclusion 3:
From V and VI, we can conclude this:
• Product of two rational numbers will always be a rational number.
• But product of two irrational numbers may be rational or irrational
Conclusion 4:
From VII and VIII, we can conclude this:
• Quotient of two rational numbers will always be a rational number.
• But quotient of two irrational numbers may be rational or irrational
Conclusion 5
• Sum of a rational number and an irrational number will always be an irrational number.
Conclusion 6
• Difference of a rational number and an irrational number will always be an irrational number.
Conclusion 7
• Product of a rational number and an irrational number will always be an irrational number.
♦ In this case it is important to mention that, the rational number should not be zero. Other wise, the product will be zero.
Conclusion 8
• Quotient of a rational number and an irrational number will always be an irrational number.
♦ In this case it is important to mention that, the rational number should not be zero. Other wise, the quotient will be zero.
The reader may write his/her own examples related to the above conclusions 5, 6, 7 and 8
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