In the previous section we completed the discussion on multiplication and division of fractions. We also saw a solved example. In this section, we will see more solved examples.
Solved example 5.39
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. One is the reciprocal of the other. But there is more..
The denominator of the first fraction is equal to (numerator +1)
2. So, we can denote the first fraction as a⁄(a+1) . The second fraction will be (a+1)⁄a
3. So the portion on the left side of the first '=' sign can be represented as: a⁄(a+1) + (a+1)⁄a
■ Consider the portion in between the two '=' signs
4. There we have a single fraction
5. It is nothing but the sum in (3). We know that a⁄(a+1) + (a+1)⁄a = [a2 + (a+1)2]⁄[a(a+1)]
■ Consider the portion to the right of the last '=' sign
6. There we have a fraction added to '2'
7. The numerator of this fraction is always '1'
8. The denominator of the fraction is the 'product of the denominators in (3)
9. So we can represent the portion to the right of the last '=' sign as: 2 + 1⁄[a×(a+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:
a⁄(a+1) + (a+1)⁄a = [a2 + (a+1)2]⁄[a(a+1)]
2. This is a simple case of addition. Addition of two fractions which are reciprocals of each other. And also denominator = (numerator +1). In fact it is the expression which can be used to add any such reciprocal fractions. We can easily derive it, and there is nothing special to prove
3. Now we prove the equality of the portions on either sides of the last '=' sign. That is ., we prove:
[a2 + (a+1)2]⁄[a(a+1)] = 2 + 1⁄[a×(a+1)]
4. The right side of (3) can be written as [2a(a+1) +1]⁄[a(a+1)]
5. Now the denominators in (3) becomes equal. All we have to prove now is that numerators are also equal. That is., we have to prove: [a2 + (a+1)2] = [2a(a+1) + 1]
6. • Expanding the left side of (5) we get: a2 + a2 + 2a +1 = 2a2 + 2a + 1
• Expanding the right side of (5) we get: 2a2 + 2a + 1 (same as above)
• So the equality in (5) is proved. Hence the equality in (3) is established
7. In the above steps, first we proved A a⁄(a+1) + (a+1)⁄a = B [a2 + (a+1)2]⁄[a(a+1)]
8. Then we proved B [a2 + (a+1)2]⁄[a(a+1)] = C 2 + 1⁄[a×(a+1)]
9. That means A = B = C
10. That is., a⁄(a+1) + (a+1)⁄a = [a2 + (a+1)2]⁄[a(a+1)] = 2 + 1⁄[a×(a+1)]
Solved example 5.40
Examine the pattern given below and write the principle behind it. Also give the algebraic proof
Solution:
■ Consider the portion inside the green box.
1. There are two unit fractions (numerators '1')
2. The second denominator = (first denominator +1)
3. The two fractions are added, and the sum is noted down
4. So this portion can be represented as 1⁄a + 1⁄(a+1)
■ Consider the portion inside the blue box
5. Here also there are two fractions. They are closely related to the first set in the green box.
6. The denominator of the second becomes the numerator of the first
7. The denominator of the first becomes numerator of the second
8. There is no change in denominators
9. Also in the blue box, the second fraction is subtracted from the first fraction
10. So the portion can be represented as: (a+1)⁄a - a⁄(a+1)
11. We can see that the results in the green box and blue box are the same
■ 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 write the expression for the left side. It is simple addition:
1⁄a + 1⁄(a+1) = [(a+1) + a]⁄[a×(a+1)]
2. Now we write the expression for the right side. It is simple subtraction:
(a+1)⁄a - a⁄(a+1) = [(a+1)2 - a2]⁄[a×(a+1)]
3. The end results in (1) and (2) must be equal. That is.,
4. We must prove: [(a+1) + a]⁄[a×(a+1)] = [(a+1)2 - a2]⁄[a×(a+1)]
5. The denominators are equal. So, we need to prove that the numerators are also equal. That is.,
6. We must prove: (a+1) +a = (a+1)2- a2
7. • Expanding the left side of (6) we get: 2a + 1
• Expanding the right side of (6) we get: a2 + 2a + 1 - a2 = 2a + 1(same as above)
• So the equality in (6) is proved. Hence the equality in (4) is established
Solved example 5.41
Examine the pattern given below and write the principle behind it. Also give the algebraic proof
Solution:
■ Consider the portion inside the green box
1. It has two mixed fractions. Any mixed fraction is equal to: (a number) + (a fraction)
2. Consider the first mixed fraction. If we denote the number part as 'a', then:
♦ The denominator of the fraction part will be (a-2)
♦ That is., the fraction part will be 1⁄(a-2)
3. So the first mixed fraction will be a + 1⁄(a-2)
4. Now consider the second mixed fraction
5. It's number part is always 1
6. It's fraction part is same as the fraction part of the 'first mixed fraction'
7. So the second mixed fraction will be 1 + 1⁄(a-2)
■ Consider the portion inside the blue box
8. Here the mixed fractions are same as those in the green box
9. The first mixed fraction is divided by the second
10. The results in the green box and blue box are the same
11. Based on the above details, we can write:
We must write the proof. It can be done as follows:
1. Consider the green box first. There are two terms. 1⁄(a-2) is common in both the terms. Let us put this common term equal to 'K'.
2. Then the portion in the green box becomes (a + K ) - (1 + K)
This is equal to a + K - 1 - K = a - 1
3.Now consider the blue box. The first term is a + 1⁄(a-2). This can be expanded:
4. a + 1⁄(a-2) = (a2-2a+1)⁄(a-2)
5. Look at the numerator (a2 -2a +1). It is the expanded form of (a-1)2
6. We can substitute this in (4). Thus we get a + 1⁄(a-2) = (a-1)2⁄(a-2)
7. Now consider the second term: 1 + 1⁄(a-2) . This can also be expanded
8. 1 + 1⁄(a-2) = (a-2+1)⁄(a-2) = (a-1)⁄(a-2)
9. So, from (6), the first term in the blue box is (a-1)2⁄(a-2) and, from (8), the second term is (a-1)⁄(a-2) 10. We have to divide the first term by the second term. It is same as multiplying the first term by the reciprocal of the second term: (a-1)2⁄(a-2) × (a-2)⁄(a-1) = (a-1)
11. So the result in the blue box is (a-1). From (2), the result in the green box is also (a-1)
12. Thus the equality is proved
We have completed the discussion on fractions. In the next chapter, we will see Decimals. Discussion on the fractions which give recurring decimals can be seen here.
Solved example 5.39
Examine the pattern given below and write the principle behind it. Also give the algebraic proof
■ Consider the portion on the left side of the first '=' sign
1. There we have two fractions. One is the reciprocal of the other. But there is more..
The denominator of the first fraction is equal to (numerator +1)
2. So, we can denote the first fraction as a⁄(a+1) . The second fraction will be (a+1)⁄a
3. So the portion on the left side of the first '=' sign can be represented as: a⁄(a+1) + (a+1)⁄a
■ Consider the portion in between the two '=' signs
4. There we have a single fraction
5. It is nothing but the sum in (3). We know that a⁄(a+1) + (a+1)⁄a = [a2 + (a+1)2]⁄[a(a+1)]
■ Consider the portion to the right of the last '=' sign
6. There we have a fraction added to '2'
7. The numerator of this fraction is always '1'
8. The denominator of the fraction is the 'product of the denominators in (3)
9. So we can represent the portion to the right of the last '=' sign as: 2 + 1⁄[a×(a+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:
a⁄(a+1) + (a+1)⁄a = [a2 + (a+1)2]⁄[a(a+1)]
2. This is a simple case of addition. Addition of two fractions which are reciprocals of each other. And also denominator = (numerator +1). In fact it is the expression which can be used to add any such reciprocal fractions. We can easily derive it, and there is nothing special to prove
3. Now we prove the equality of the portions on either sides of the last '=' sign. That is ., we prove:
[a2 + (a+1)2]⁄[a(a+1)] = 2 + 1⁄[a×(a+1)]
4. The right side of (3) can be written as [2a(a+1) +1]⁄[a(a+1)]
5. Now the denominators in (3) becomes equal. All we have to prove now is that numerators are also equal. That is., we have to prove: [a2 + (a+1)2] = [2a(a+1) + 1]
6. • Expanding the left side of (5) we get: a2 + a2 + 2a +1 = 2a2 + 2a + 1
• Expanding the right side of (5) we get: 2a2 + 2a + 1 (same as above)
• So the equality in (5) is proved. Hence the equality in (3) is established
7. In the above steps, first we proved A a⁄(a+1) + (a+1)⁄a = B [a2 + (a+1)2]⁄[a(a+1)]
8. Then we proved B [a2 + (a+1)2]⁄[a(a+1)] = C 2 + 1⁄[a×(a+1)]
9. That means A = B = C
10. That is., a⁄(a+1) + (a+1)⁄a = [a2 + (a+1)2]⁄[a(a+1)] = 2 + 1⁄[a×(a+1)]
Solved example 5.40
Examine the pattern given below and write the principle behind it. Also give the algebraic proof
■ Consider the portion inside the green box.
1. There are two unit fractions (numerators '1')
2. The second denominator = (first denominator +1)
3. The two fractions are added, and the sum is noted down
4. So this portion can be represented as 1⁄a + 1⁄(a+1)
■ Consider the portion inside the blue box
5. Here also there are two fractions. They are closely related to the first set in the green box.
6. The denominator of the second becomes the numerator of the first
7. The denominator of the first becomes numerator of the second
8. There is no change in denominators
9. Also in the blue box, the second fraction is subtracted from the first fraction
10. So the portion can be represented as: (a+1)⁄a - a⁄(a+1)
11. We can see that the results in the green box and blue box are the same
■ Based on the above details, we can represent the given pattern algebraically as:
1. First we write the expression for the left side. It is simple addition:
1⁄a + 1⁄(a+1) = [(a+1) + a]⁄[a×(a+1)]
2. Now we write the expression for the right side. It is simple subtraction:
(a+1)⁄a - a⁄(a+1) = [(a+1)2 - a2]⁄[a×(a+1)]
3. The end results in (1) and (2) must be equal. That is.,
4. We must prove: [(a+1) + a]⁄[a×(a+1)] = [(a+1)2 - a2]⁄[a×(a+1)]
5. The denominators are equal. So, we need to prove that the numerators are also equal. That is.,
6. We must prove: (a+1) +a = (a+1)2- a2
7. • Expanding the left side of (6) we get: 2a + 1
• Expanding the right side of (6) we get: a2 + 2a + 1 - a2 = 2a + 1(same as above)
• So the equality in (6) is proved. Hence the equality in (4) is established
Solved example 5.41
Examine the pattern given below and write the principle behind it. Also give the algebraic proof
■ Consider the portion inside the green box
1. It has two mixed fractions. Any mixed fraction is equal to: (a number) + (a fraction)
2. Consider the first mixed fraction. If we denote the number part as 'a', then:
♦ The denominator of the fraction part will be (a-2)
♦ That is., the fraction part will be 1⁄(a-2)
3. So the first mixed fraction will be a + 1⁄(a-2)
4. Now consider the second mixed fraction
5. It's number part is always 1
6. It's fraction part is same as the fraction part of the 'first mixed fraction'
7. So the second mixed fraction will be 1 + 1⁄(a-2)
■ Consider the portion inside the blue box
8. Here the mixed fractions are same as those in the green box
9. The first mixed fraction is divided by the second
10. The results in the green box and blue box are the same
11. Based on the above details, we can write:
1. Consider the green box first. There are two terms. 1⁄(a-2) is common in both the terms. Let us put this common term equal to 'K'.
2. Then the portion in the green box becomes (a + K ) - (1 + K)
This is equal to a + K - 1 - K = a - 1
3.Now consider the blue box. The first term is a + 1⁄(a-2). This can be expanded:
4. a + 1⁄(a-2) = (a2-2a+1)⁄(a-2)
5. Look at the numerator (a2 -2a +1). It is the expanded form of (a-1)2
6. We can substitute this in (4). Thus we get a + 1⁄(a-2) = (a-1)2⁄(a-2)
7. Now consider the second term: 1 + 1⁄(a-2) . This can also be expanded
8. 1 + 1⁄(a-2) = (a-2+1)⁄(a-2) = (a-1)⁄(a-2)
9. So, from (6), the first term in the blue box is (a-1)2⁄(a-2) and, from (8), the second term is (a-1)⁄(a-2) 10. We have to divide the first term by the second term. It is same as multiplying the first term by the reciprocal of the second term: (a-1)2⁄(a-2) × (a-2)⁄(a-1) = (a-1)
11. So the result in the blue box is (a-1). From (2), the result in the green box is also (a-1)
12. Thus the equality is proved
We have completed the discussion on fractions. In the next chapter, we will see Decimals. Discussion on the fractions which give recurring decimals can be seen here.
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