Chapter 5.11 - Equivalent fractions and Cross multiplication

In the previous section we learned about subtraction of fractions. In this section we will see some advanced features of fractions.

We have already seen equivalent fractions. Some examples are:
• ¹⁄₂ = ²⁄₄ = ³⁄₆ = . . . .
    ♦ We can say that ²⁄₄ , ³⁄₆ , . . . . are all different forms of ¹⁄₂
    ♦ We can also say that ²⁄₄ , ³⁄₆ , . . . . are all equal to ¹⁄₂  
• ³⁄₅ = ⁶⁄₁₀ = ⁹⁄₁₅ = . . . .
    ♦ We can say that ⁶⁄₁₀ , ⁹⁄₁₅ , . . . . are all different forms of ³⁄₅
    ♦ We can also say that ⁶⁄₁₀ , ⁹⁄₁₅ , . . . . are all equal to ³⁄₅  
When we multiply both the numerator and denominator of a fraction by the same number, we get different forms of that fraction. The new fractions thus obtained will be equal in value to the original  fraction.

We can put this in algebra:

Now consider the following pattern:

It can be explained as follows:
• In the numerator: A number is added to it's square 
• In the denominator: The number is added to 1
• The result is the number itself.

Will we get the same result for all numbers? For example:

Is the result equal to 127 ? We can use a calculator. But it is better to use another method. It will enable us to know the reason for getting a particular answer. For that, we rewrite the numerator:
127² + 127 = 127 (127 +1)
Now we can rewrite the expression as

• In the above step, both numerator and denominator of  ¹²⁷⁄₁ is multiplied by (127 +1). 
• So it has no effect. We get ¹²⁷⁄₁ itself. 
• But ¹²⁷⁄₁ is equal to 127. Thus we get the original number. 
• We can write this using algebra:

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Now we will see some solved examples
Solved example 5.33
Explain each of the patterns below, and write the general principle in algebra.

Solution:
1 • In the numerator: A number is added to it's square 
• In the denominator: The number is added to itself
• The result is: half of (number +1)
• We can prove it using algebra: 

• In the above step, n² + n is written as n(n+1)
• So the numerator and denominator of (n+1)/2 is multiplied by n
• Thus we get (n+1)/2 itself.

2 • In the numerator: A number is subtracted from it's square 
• In the denominator:  1 is subtracted from the number
• The result is:  The number itself
• We can prove this using algebra:

 • In the above step,  n² - n is written as n(n-1)
• So the numerator and denominator of n/1 is multiplied by (n-1)
• So we get n/1 itself. But n/1 = n

3 • In the numerator: 1 is subtracted from the square of a number 
• In the denominator:  1 is subtracted from the number
• The result is: (number + 1)
• We can prove this using algebra:

• In the above step,  n² - 1 is written as (n+1)(n-1) [Using identities: (a+b)(a-b) = a² - b²]
• So the numerator and denominator of  ⁽ⁿ⁺¹⁾⁄₁ is multiplied by (n-1)

• So we get ⁽ⁿ⁺¹⁾⁄₁ itself. But ⁽ⁿ⁺¹⁾⁄₁ = n +1

Cross Multiplication
We have seen the basics about cross multiplication here. We used it to find equivalent fractions. Now we will give it an algebraic form:
• a/b and p/q are two given fractions. We want to know whether they are equivalent. 
• For that, we multiply both the numerator and denominator of a/b by q. This will give us aq/bq
• Next we multiply both the numerator and denominator of p/q by b. This will give us pb/qb
• From the above 2 steps, we get two new fractions: aq/bq and pb/qb The denominators of both these fractions are equal. (since bq = qb)

• If the numerators are also equal, we can conclude that the original fractions a/b and p/q are equivalent.
The above steps can be written as:

We can write the converse also:

This is simple multiplication of the numerator of one fraction with the denominator of the other.

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In the next section we will discuss more about fractions.

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