In the previous section we completed the discussion on addition and subtraction of fractions. In this section, we will see multiplication and division.
We know that, to multiply two fractions, we multiply the numerators and the denominators separately. An example is given below:
Using algebra, the multiplication can be written as:
We also know division: We multiply the reciprocal of the fraction. An example is given below:
Using algebra, the division can be written as:
Consider the following pattern:
The pattern can be explained as follows:
■ We have two fractions 1⁄2 and 1⁄2 with us. They can be any two fractions. But the condition is that, their sum must be equal to 1
• In step 1, the reciprocals of the original fractions are added. The sum thus obtained is 4
• In step 2, the reciprocals are multiplied. The product thus obtained is 4
■ We have two fractions 1⁄3 and 2⁄3 with us. Their sum is equal to 1
• In step 1, the reciprocals of the original fractions are added. The sum thus obtained is 9⁄2
• In step 2, the reciprocals are multiplied. The product thus obtained is 9⁄2
■ We have two fractions 2⁄9 and 7⁄9 with us. Their sum is equal to 1
• In step 1, the reciprocals of the original fractions are added. The sum thus obtained is 81⁄14
• In step 2, the reciprocals are multiplied. The product thus obtained is 81⁄14
Based on the above, we can write a general form:
But we have to show proof. it can be done as follows:
1. Let us start with a⁄b + p⁄q = 1
2. From this we get (aq +pb)⁄bq =1 ⇒ aq + pb = bq
3. Now take b⁄a + q⁄p
4. From this we get b⁄a + q⁄p = (bp +qa)⁄pa
5. But from (3), bp + qa = bq. Putting this in (4), we get:
6. b⁄a + q⁄p = bq⁄pa
7. Now take b⁄a × q⁄p . This is same as bq⁄pa
8. That is., b⁄a × q⁄p = bq⁄pa
9. Comparing (6) and (8), we get: b⁄a + q⁄p = bq⁄pa
10. Hence proved
We will now see a solved example
Solved example 5.38
Examine the pattern given below and write the principle behind it. Also give the algebraic proof
Solution:
■ Consider the portion on the left side of the first '=' sign
1. There we have two fractions. The numerators of both are '1'. So the fractions that we are having are 'unit fractions'
2. The denominators are alternate numbers. So, if we denote the first denominator as 'a', then the second denominator will be 'a+2'
3. So the portions on the left side of the first '=' sign can be represented as: 1⁄a - 1⁄(a+2)
■ Consider the portion in between the two '=' signs
4. There we have a single fraction
5. The numerator is always '2'
6. The denominator is the product of the denominators in (3)
7. So, the portion between the two '=' signs can be represented as: 2⁄[a×(a+2)]
■ Consider the portion to the right of the last '=' sign
8. There we have a single fraction
9. The numerator is always '2'
10. The denominator can be represented as [(a+1)2 - 1]
♦ Why (a+1)2 ?
♦ Because it is the square of 'the number in between the two denominators in (3)': 'a' and 'a+2'
♦ The number in between 'a' and 'a+2' is 'a+1'
11. So the portion to the right of the last '=' sign can be represented as: 2⁄[(a+1)2-1]
■ Based on the above details, we can represent the given pattern algebraically as:
We must write the proof. It can be done as follows:
1. First we prove the equality of the portions on either sides of the first '=' sign. That is ., we prove: 1⁄a - 1⁄(a+2) = 2⁄[a×(a+2)]
2. The left side of the above, is a simple case. It is the subtraction of a unit fraction from another unit fraction. We have seen it here.
3. So we can write: 1⁄a - 1⁄(a+2) = [(a+2) - a]⁄[a×(a+2)] = 2⁄[a×(a+2)] (∵ a and -a cancel each other)
4. Thus the equality in (1) is proved
5. Now we prove the equality of the portions on either sides of the last '=' sign. That is ., we prove: 2⁄[a×(a+2)] = 2⁄[(a+1)2-1]
6. Here the numerators are equal. So we need to prove that the denominators are also equal
7. That is., we need to prove only this: a×(a+2) = (a+1)2 - 1
8. Expanding the left side, we get: a2 + 2a
9. Expanding the right side, we get: a2 + 2a + 1 - 1 = a2 + 2a
10. (8) is equal to (9). So (7) is proved. And consequently, (5) is proved
11. In the above steps, first we proved A 1⁄a - 1⁄(a+2) = B 2⁄[a×(a+2)]
12. Then we proved B 2⁄[a×(a+2)] = C 2⁄[(a+1)2-1]
13. That means A = B = C
14. That is., 1⁄a - 1⁄(a+2) = 2⁄[a×(a+2)] = 2⁄[(a+1)2-1]
In the next section we will see more solved examples.
We know that, to multiply two fractions, we multiply the numerators and the denominators separately. An example is given below:
Using algebra, the multiplication can be written as:
We also know division: We multiply the reciprocal of the fraction. An example is given below:
Using algebra, the division can be written as:
Consider the following pattern:
■ We have two fractions 1⁄2 and 1⁄2 with us. They can be any two fractions. But the condition is that, their sum must be equal to 1
• In step 1, the reciprocals of the original fractions are added. The sum thus obtained is 4
• In step 2, the reciprocals are multiplied. The product thus obtained is 4
■ We have two fractions 1⁄3 and 2⁄3 with us. Their sum is equal to 1
• In step 1, the reciprocals of the original fractions are added. The sum thus obtained is 9⁄2
• In step 2, the reciprocals are multiplied. The product thus obtained is 9⁄2
■ We have two fractions 2⁄9 and 7⁄9 with us. Their sum is equal to 1
• In step 1, the reciprocals of the original fractions are added. The sum thus obtained is 81⁄14
• In step 2, the reciprocals are multiplied. The product thus obtained is 81⁄14
Based on the above, we can write a general form:
But we have to show proof. it can be done as follows:
1. Let us start with a⁄b + p⁄q = 1
2. From this we get (aq +pb)⁄bq =1 ⇒ aq + pb = bq
3. Now take b⁄a + q⁄p
4. From this we get b⁄a + q⁄p = (bp +qa)⁄pa
5. But from (3), bp + qa = bq. Putting this in (4), we get:
6. b⁄a + q⁄p = bq⁄pa
7. Now take b⁄a × q⁄p . This is same as bq⁄pa
8. That is., b⁄a × q⁄p = bq⁄pa
9. Comparing (6) and (8), we get: b⁄a + q⁄p = bq⁄pa
10. Hence proved
We will now see a solved example
Solved example 5.38
Examine the pattern given below and write the principle behind it. Also give the algebraic proof
Solution:
■ Consider the portion on the left side of the first '=' sign
1. There we have two fractions. The numerators of both are '1'. So the fractions that we are having are 'unit fractions'
2. The denominators are alternate numbers. So, if we denote the first denominator as 'a', then the second denominator will be 'a+2'
3. So the portions on the left side of the first '=' sign can be represented as: 1⁄a - 1⁄(a+2)
■ Consider the portion in between the two '=' signs
4. There we have a single fraction
5. The numerator is always '2'
6. The denominator is the product of the denominators in (3)
7. So, the portion between the two '=' signs can be represented as: 2⁄[a×(a+2)]
■ Consider the portion to the right of the last '=' sign
8. There we have a single fraction
9. The numerator is always '2'
10. The denominator can be represented as [(a+1)2 - 1]
♦ Why (a+1)2 ?
♦ Because it is the square of 'the number in between the two denominators in (3)': 'a' and 'a+2'
♦ The number in between 'a' and 'a+2' is 'a+1'
11. So the portion to the right of the last '=' sign can be represented as: 2⁄[(a+1)2-1]
■ Based on the above details, we can represent the given pattern algebraically as:
1. First we prove the equality of the portions on either sides of the first '=' sign. That is ., we prove: 1⁄a - 1⁄(a+2) = 2⁄[a×(a+2)]
2. The left side of the above, is a simple case. It is the subtraction of a unit fraction from another unit fraction. We have seen it here.
3. So we can write: 1⁄a - 1⁄(a+2) = [(a+2) - a]⁄[a×(a+2)] = 2⁄[a×(a+2)] (∵ a and -a cancel each other)
4. Thus the equality in (1) is proved
5. Now we prove the equality of the portions on either sides of the last '=' sign. That is ., we prove: 2⁄[a×(a+2)] = 2⁄[(a+1)2-1]
6. Here the numerators are equal. So we need to prove that the denominators are also equal
7. That is., we need to prove only this: a×(a+2) = (a+1)2 - 1
8. Expanding the left side, we get: a2 + 2a
9. Expanding the right side, we get: a2 + 2a + 1 - 1 = a2 + 2a
10. (8) is equal to (9). So (7) is proved. And consequently, (5) is proved
11. In the above steps, first we proved A 1⁄a - 1⁄(a+2) = B 2⁄[a×(a+2)]
12. Then we proved B 2⁄[a×(a+2)] = C 2⁄[(a+1)2-1]
13. That means A = B = C
14. That is., 1⁄a - 1⁄(a+2) = 2⁄[a×(a+2)] = 2⁄[(a+1)2-1]
In the next section we will see more solved examples.
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