In the previous section we saw the comparison of unequal fractions. Now we will see a more advanced case.
Consider the following example:
• We have a fraction 2⁄3 in hand
• We are going to make a new fraction from this 2⁄3
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is (2+1)⁄(3+1) = 3⁄4
• Now compare the new fraction 3⁄4 with the original fraction 2⁄3
■ We find that the new fraction is greater than the original fraction
Another example:
• We have a fraction 5⁄9 in hand
• We are going to make a new fraction from this 5⁄9
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is (5+1)⁄(9+1) = 6⁄10
• Now compare the new fraction 6⁄10 with the original fraction 5⁄9
■ We find that the new fraction is greater than the original fraction
We find that the new fraction is greater than the original fraction in both the above cases. But is this always true? Let us see another example:
• We have a fraction 4⁄3 in hand
• We are going to make a new fraction from this 4⁄3
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is (4+1)⁄(3+1) = 5⁄4
• Now compare the new fraction 5⁄4 with the original fraction 4⁄3
In this example, we see that the new fraction is less than the original fraction. Why is this so?
• In the previous two examples, the numerator of the original fractions 2⁄3 and 5⁄9 is less than the denominator
• In the present example, the numerator of the original fraction 4⁄3 is greater than the denominator
So we can write it algebraically:
■ If a < b, then a⁄b < (a+1)⁄(b+1)
■ If a > b, then (a+1)⁄(b+1) < a⁄b
But we have to show proof for the above. It can be done as follows:
1. If a < b, then prove that a⁄b < (a+1)⁄(b+1)
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that a(b+1) < b(a+1)
4. Expanding the left side of (3), we get: ab + a
5. Expanding the right side of (3) we get: ba + b
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
8. It is given in (1) that a < b. So we find that left side of (3) is indeed inferior.
9. Hence (3) is proved, and consequently, (1) is established
Similarly we can prove the second case also as follows:
1. If a > b, then prove that (a+1)⁄(b+1) < a⁄b
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that (a+1)b < a(b+1)
4. Expanding the left side of (3), we get: ab + b
5. Expanding the right side of (3) we get: ba + a
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
8. It is given in (1) that a > b. So we find that left side of (3) is indeed inferior.
9. Hence (3) is proved, and consequently, (1) is established
We will now see some solved examples
Solved example 5.36
Find the larger of each pair of fractions below with out pen and paper
(i) 3⁄5 , 8⁄13 (ii) 3⁄5 ,6⁄11 (iii) 74⁄75, 75⁄76 (iv) 101⁄102 , 98⁄99
Solution:
(i) We can use the property: if aq < pb then, a⁄b < p⁄q (Details here)
1. In the given problem, aq = 3 × 13 = 39, and pb = 8 × 5 = 40
2. 39 < 40. So aq < pb.
3. Thus we get a⁄b < p⁄q ⇒ 3⁄5 < 8⁄13
Once we understand the basic principles, we can do the above steps with out pen and paper
(ii) We can use the property: if aq < pb then, a⁄b < p⁄q
1. In the given problem, aq = 3 × 11 = 33, and pb = 6 × 5 = 30
2. 30 < 33. So pb < aq.
3. Thus we get p⁄q < a⁄b ⇒ 6⁄11 < 3⁄5
Once we understand the basic principles, we can do the above steps with out pen and paper
(iii) We can use the property that we saw earlier in this section: If a < b, then a⁄b < (a+1)⁄(b+1)
1. We have a⁄b = 74⁄75 and (a+1)⁄(b+1) = 75⁄76
2. So we get 74⁄75 < 75⁄76
(iv) We can use the property that we saw earlier in this section: If a < b, then a⁄b < (a+1)⁄(b+1)
1. Take a⁄b = 98⁄99 So that, (a+1)⁄(b+1) = 99⁄100
2. From this we get 98⁄99 < 99⁄100
3. Take a⁄b = 99⁄100 So that, (a+1)⁄(b+1) = 100⁄101
4. From this we get 99⁄100 < 100⁄101
5. Take a ⁄b = 100⁄101 So that, (a+1)⁄(b+1) = 101⁄102
6. From this we get 100⁄101 < 101⁄102
7. Consider the results in (2), (4) and (6) together: 98⁄99 < 99⁄100 < 100⁄101 < 101⁄102
8. From the first and last terms of the result in (7), we get 98⁄99 < 101⁄102
Once we understand the basic principles, we can do the above steps with out pen and paper
In the property that we discussed above, we increased the numerator and denominator by '1'. What if we increase by other natural numbers 2, 3, 4.. etc.,? Do we get the same result? We can check using algebra. We can put 'n' to represent any natural number.
1. If a < b, then prove that a⁄b < (a+n)⁄(b+n)
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that a(b+n) < b(a+n)
4. Expanding the left side of (3), we get: ab + an
5. Expanding the right side of (3) we get: ba + bn
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by an and bn
8. Here also 'n' is common. It will cancel each other
9. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
10. It is given in (1) that a < b. So we find that left side of (3) is indeed inferior
11. Hence (3) is proved, and consequently, (1) is established
Similarly we can prove the second case also as follows:
1. If a > b, then prove that (a+n)⁄(b+n) < a⁄b
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that (a+n)b < a(b+n)
4. Expanding the left side of (3), we get: ab + nb
5. Expanding the right side of (3) we get: ba + na
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by nb and na
8. Here also 'n' is common. It will cancel each other
9. So the superiority or inferiority of the left side and right side in (3) is decided by b and a
10. It is given in (1) that a > b. So we find that left side of (3) is indeed inferior
11. Hence (3) is proved, and consequently, (1) is established
In the next section we will see an even more advanced case.
Consider the following example:
• We have a fraction 2⁄3 in hand
• We are going to make a new fraction from this 2⁄3
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is (2+1)⁄(3+1) = 3⁄4
• Now compare the new fraction 3⁄4 with the original fraction 2⁄3
Another example:
• We have a fraction 5⁄9 in hand
• We are going to make a new fraction from this 5⁄9
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is (5+1)⁄(9+1) = 6⁄10
• Now compare the new fraction 6⁄10 with the original fraction 5⁄9
■ We find that the new fraction is greater than the original fraction
We find that the new fraction is greater than the original fraction in both the above cases. But is this always true? Let us see another example:
• We have a fraction 4⁄3 in hand
• We are going to make a new fraction from this 4⁄3
• For that, we increase both the numerator and denominator by 1
• So the resulting fraction is (4+1)⁄(3+1) = 5⁄4
• Now compare the new fraction 5⁄4 with the original fraction 4⁄3
• In the previous two examples, the numerator of the original fractions 2⁄3 and 5⁄9 is less than the denominator
• In the present example, the numerator of the original fraction 4⁄3 is greater than the denominator
So we can write it algebraically:
■ If a < b, then a⁄b < (a+1)⁄(b+1)
■ If a > b, then (a+1)⁄(b+1) < a⁄b
But we have to show proof for the above. It can be done as follows:
1. If a < b, then prove that a⁄b < (a+1)⁄(b+1)
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that a(b+1) < b(a+1)
4. Expanding the left side of (3), we get: ab + a
5. Expanding the right side of (3) we get: ba + b
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
8. It is given in (1) that a < b. So we find that left side of (3) is indeed inferior.
9. Hence (3) is proved, and consequently, (1) is established
Similarly we can prove the second case also as follows:
1. If a > b, then prove that (a+1)⁄(b+1) < a⁄b
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that (a+1)b < a(b+1)
4. Expanding the left side of (3), we get: ab + b
5. Expanding the right side of (3) we get: ba + a
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
8. It is given in (1) that a > b. So we find that left side of (3) is indeed inferior.
9. Hence (3) is proved, and consequently, (1) is established
We will now see some solved examples
Solved example 5.36
Find the larger of each pair of fractions below with out pen and paper
(i) 3⁄5 , 8⁄13 (ii) 3⁄5 ,6⁄11 (iii) 74⁄75, 75⁄76 (iv) 101⁄102 , 98⁄99
Solution:
(i) We can use the property: if aq < pb then, a⁄b < p⁄q (Details here)
1. In the given problem, aq = 3 × 13 = 39, and pb = 8 × 5 = 40
2. 39 < 40. So aq < pb.
3. Thus we get a⁄b < p⁄q ⇒ 3⁄5 < 8⁄13
Once we understand the basic principles, we can do the above steps with out pen and paper
(ii) We can use the property: if aq < pb then, a⁄b < p⁄q
1. In the given problem, aq = 3 × 11 = 33, and pb = 6 × 5 = 30
2. 30 < 33. So pb < aq.
3. Thus we get p⁄q < a⁄b ⇒ 6⁄11 < 3⁄5
Once we understand the basic principles, we can do the above steps with out pen and paper
(iii) We can use the property that we saw earlier in this section: If a < b, then a⁄b < (a+1)⁄(b+1)
1. We have a⁄b = 74⁄75 and (a+1)⁄(b+1) = 75⁄76
2. So we get 74⁄75 < 75⁄76
(iv) We can use the property that we saw earlier in this section: If a < b, then a⁄b < (a+1)⁄(b+1)
1. Take a⁄b = 98⁄99 So that, (a+1)⁄(b+1) = 99⁄100
2. From this we get 98⁄99 < 99⁄100
3. Take a⁄b = 99⁄100 So that, (a+1)⁄(b+1) = 100⁄101
4. From this we get 99⁄100 < 100⁄101
5. Take a ⁄b = 100⁄101 So that, (a+1)⁄(b+1) = 101⁄102
6. From this we get 100⁄101 < 101⁄102
7. Consider the results in (2), (4) and (6) together: 98⁄99 < 99⁄100 < 100⁄101 < 101⁄102
8. From the first and last terms of the result in (7), we get 98⁄99 < 101⁄102
Once we understand the basic principles, we can do the above steps with out pen and paper
In the property that we discussed above, we increased the numerator and denominator by '1'. What if we increase by other natural numbers 2, 3, 4.. etc.,? Do we get the same result? We can check using algebra. We can put 'n' to represent any natural number.
1. If a < b, then prove that a⁄b < (a+n)⁄(b+n)
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that a(b+n) < b(a+n)
4. Expanding the left side of (3), we get: ab + an
5. Expanding the right side of (3) we get: ba + bn
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by an and bn
8. Here also 'n' is common. It will cancel each other
9. So the superiority or inferiority of the left side and right side in (3) is decided by a and b
10. It is given in (1) that a < b. So we find that left side of (3) is indeed inferior
11. Hence (3) is proved, and consequently, (1) is established
Similarly we can prove the second case also as follows:
1. If a > b, then prove that (a+n)⁄(b+n) < a⁄b
2. We have seen earlier that: if a⁄b < p⁄q then, aq < pb
3. So to prove (1), we need to prove that (a+n)b < a(b+n)
4. Expanding the left side of (3), we get: ab + nb
5. Expanding the right side of (3) we get: ba + na
6. ab is same as ba. That means we have 'one term same' on both sides in (3). They will cancel out each other
7. So the superiority or inferiority of the left side and right side in (3) is decided by nb and na
8. Here also 'n' is common. It will cancel each other
9. So the superiority or inferiority of the left side and right side in (3) is decided by b and a
10. It is given in (1) that a > b. So we find that left side of (3) is indeed inferior
11. Hence (3) is proved, and consequently, (1) is established
In the next section we will see an even more advanced case.
No comments:
Post a Comment