In the previous section we completed the discussion on the various properties of fractions. In this section, we will see addition and subtraction of fractions.
We know that,when the denominators are equal, the addition of fractions is easy. The denominator of the result will be the denominator in the given problem. The numerator will be the sum of the numerators.
When adding fractions with different denominators, more work is involved. We have already seen the process here. Let us see an example:
In the above process,
• We multiplied the numerator and denominator of first fraction 1⁄2 with the 'other denominator', which is '3'
• We multiplied the numerator and denominator of the second fraction 1⁄3 also with the 'other denominator', which is '2'
• Thus we get two fractions with the same denominators
We can write the process in algebraic form:
Based on this, we can write the general form:
For subtraction, the general form will be:
Example for addition:
Example for subtraction:
The above operations will become very easy if the numerators are all '1'. Let us check such cases algebraically:
So we can write the general form for addition of fractions with numerator '1' as:
And the general form for the subtraction of fractions with numerator '1' can be written as:
Addition example:
Subtraction example:
Fractions with Numerator '1' are called unit fractions. We will now see some properties of unit fractions:
Consider the pattern:
We can continue the steps any number of times. But for our present discussion, the few steps written above are sufficient.
1. Consider the sum: [1⁄1 - 1⁄2] + [1⁄2 - 1⁄3 ]. What is the result?
2. -1⁄2 and 1⁄2 will cancel out each other. So the sum is 1⁄1 - 1⁄3
3. But from the above pattern, we have: [1⁄1 - 1⁄2] = 1⁄(1×2) and [1⁄2 - 1⁄3 ] = 1⁄(2×3)
4. So comparing (2) and (3), we get: 1⁄(1×2) + 1⁄(2×3) = 1⁄1 - 1⁄3
Next, let us take three terms:
1. we get [1⁄1 - 1⁄2] + [1⁄2 - 1⁄3 ] + [1⁄3 - 1⁄4 ]. What is the result?
2. -1⁄2 and 1⁄2 will cancel out each other. Also -1⁄3 and 1⁄3 will cancel out each other. So the sum is 1⁄1 - 1⁄4
3. But from the pattern, we have:
[1⁄1 - 1⁄2] = 1⁄(1×2) ,
[1⁄2 - 1⁄3 ] = 1⁄(2×3) and
[1⁄2 - 1⁄3 ] = 1⁄(3×4)
4. So comparing (2) and (3), we get: 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4)= 1⁄1 - 1⁄4
Next, let us take four terms:
1. we get [1⁄1 - 1⁄2] + [1⁄2 - 1⁄3 ] + [1⁄3 - 1⁄4 ] + [1⁄4 - 1⁄5 ]. What is the result?
2. -1⁄2 and 1⁄2 will cancel out each other.
-1⁄3 and 1⁄3 will cancel out each other.
-1⁄4 and 1⁄4 will cancel out each other
So the sum is 1⁄1 - 1⁄5
3. But from the pattern, we have:
[1⁄1 - 1⁄2] = 1⁄(1×2) ,
[1⁄2 - 1⁄3 ] = 1⁄(2×3) ,
[1⁄2 - 1⁄3 ] = 1⁄(3×4) and
[1⁄2 - 1⁄3 ] = 1⁄(4×5)
4. So comparing (2) and (3), we get: 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4) + 1⁄(4×5)= 1⁄1 - 1⁄5
We considered three cases:
Case 1: We took two terms, and got the result:
■ 1⁄(1×2) + 1⁄(2×3) = 1⁄1 - 1⁄3
Case 2: We took three terms, and got the result:
■ 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4)= 1⁄1 - 1⁄4
Case 3: We took four terms, and got the result:
■ 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4) + 1⁄(4×5) = 1⁄1 - 1⁄5
We can go on like this and take any number of cases as we like. But that will not be necessary. A pattern has already emerged from the three cases. When we understand the pattern, we will be able to write the answer for any given case. Here is an example:
Case 98:
■ 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4) + . . . . + 1⁄(99×100) = 1⁄1 - 1⁄100 = 1 - 0.01 = 0.99
In the next section we will see multiplication and division with fractions.
We know that,when the denominators are equal, the addition of fractions is easy. The denominator of the result will be the denominator in the given problem. The numerator will be the sum of the numerators.
When adding fractions with different denominators, more work is involved. We have already seen the process here. Let us see an example:
• We multiplied the numerator and denominator of first fraction 1⁄2 with the 'other denominator', which is '3'
• We multiplied the numerator and denominator of the second fraction 1⁄3 also with the 'other denominator', which is '2'
• Thus we get two fractions with the same denominators
We can write the process in algebraic form:
Example for addition:
The above operations will become very easy if the numerators are all '1'. Let us check such cases algebraically:
Addition example:
Fractions with Numerator '1' are called unit fractions. We will now see some properties of unit fractions:
Consider the pattern:
1. Consider the sum: [1⁄1 - 1⁄2] + [1⁄2 - 1⁄3 ]. What is the result?
2. -1⁄2 and 1⁄2 will cancel out each other. So the sum is 1⁄1 - 1⁄3
3. But from the above pattern, we have: [1⁄1 - 1⁄2] = 1⁄(1×2) and [1⁄2 - 1⁄3 ] = 1⁄(2×3)
4. So comparing (2) and (3), we get: 1⁄(1×2) + 1⁄(2×3) = 1⁄1 - 1⁄3
Next, let us take three terms:
1. we get [1⁄1 - 1⁄2] + [1⁄2 - 1⁄3 ] + [1⁄3 - 1⁄4 ]. What is the result?
2. -1⁄2 and 1⁄2 will cancel out each other. Also -1⁄3 and 1⁄3 will cancel out each other. So the sum is 1⁄1 - 1⁄4
3. But from the pattern, we have:
[1⁄1 - 1⁄2] = 1⁄(1×2) ,
[1⁄2 - 1⁄3 ] = 1⁄(2×3) and
[1⁄2 - 1⁄3 ] = 1⁄(3×4)
4. So comparing (2) and (3), we get: 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4)= 1⁄1 - 1⁄4
Next, let us take four terms:
1. we get [1⁄1 - 1⁄2] + [1⁄2 - 1⁄3 ] + [1⁄3 - 1⁄4 ] + [1⁄4 - 1⁄5 ]. What is the result?
2. -1⁄2 and 1⁄2 will cancel out each other.
-1⁄3 and 1⁄3 will cancel out each other.
-1⁄4 and 1⁄4 will cancel out each other
So the sum is 1⁄1 - 1⁄5
3. But from the pattern, we have:
[1⁄1 - 1⁄2] = 1⁄(1×2) ,
[1⁄2 - 1⁄3 ] = 1⁄(2×3) ,
[1⁄2 - 1⁄3 ] = 1⁄(3×4) and
[1⁄2 - 1⁄3 ] = 1⁄(4×5)
4. So comparing (2) and (3), we get: 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4) + 1⁄(4×5)= 1⁄1 - 1⁄5
We considered three cases:
Case 1: We took two terms, and got the result:
■ 1⁄(1×2) + 1⁄(2×3) = 1⁄1 - 1⁄3
Case 2: We took three terms, and got the result:
■ 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4)= 1⁄1 - 1⁄4
Case 3: We took four terms, and got the result:
■ 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4) + 1⁄(4×5) = 1⁄1 - 1⁄5
We can go on like this and take any number of cases as we like. But that will not be necessary. A pattern has already emerged from the three cases. When we understand the pattern, we will be able to write the answer for any given case. Here is an example:
Case 98:
■ 1⁄(1×2) + 1⁄(2×3) + 1⁄(3×4) + . . . . + 1⁄(99×100) = 1⁄1 - 1⁄100 = 1 - 0.01 = 0.99
In the next section we will see multiplication and division with fractions.
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