Base equalization technique is indispensable for solving fraction problems

It is a more general technique

Base equalization technique in fraction problems

Base equalization technique produces quick solutions in indices problems. But that is not the only area of application of this more general concept. For solving problems on fractions also, the base equalization technique is indispensable.

If denominators in an expression involving fractions are different, we need to transform the individual fractional terms such that the denominators become equal. Only then we would take inverse of the denominator as a factor out of the expression for further simple manipulations of the terms now without denominator to be possible. This is how we apply base equalization technique for solving problems in fractions.

Fractions

In its simplest use, a fraction is a special way of expressing a number. $\displaystyle\frac{1}{2}$, $\displaystyle\frac{4}{5}$, $\displaystyle\frac{17}{12}$ are all fractions in numbers.

But a fraction can be more complex.

$\displaystyle\frac{p}{q}$, $\displaystyle\frac{a^2 + b^2}{a^2 - b^2}$, $\displaystyle\frac{tan\theta}{1 + sin\theta{\cos\theta}}$ also are all fractions.

In general, a fraction can be expressed as,

Fraction, $F = \displaystyle\frac{a}{b}$, where $a$ and $b$ are real numbers or expressions in variables and $b\neq{0}$.

Mathematically this is a more precise definition.

The $a$ above is universally referred to as ‘numerator’ and $b$ as ‘denominator’.

Effectively in a fraction, the numerator is divided by the denominator. But generally it is impracticable to actually carry out the division and we keep the fraction in its numerator divided by denominator form, manipulate it as required and finally simplify it to its simplest form.

Aside: The real numbers that can be expressed in $\displaystyle\frac{a}{b}$ form, where $a$ and $b$ are integers are the much favored Rational numbers. All other real numbers that cannot be expressed as a fraction of this form are lazily named as Irrational numbers. Surds such as $\sqrt{2}$ is an example of an irrational number. Evaluation of a rational number results in a decimal that will have non-terminating but recurring digit sequence or terminating digit sequence, such as $0.333333......$ or $0.675$. An irrational number is quite special in this regard as it generates a decimal number that is a non-terminating but non-recurring digit sequence such as my favorite, $.01001000100001.....$.

Use of base equalization technique in fraction arithmetic

Comparison of fractions

Frequently we need to do comparison of fractions. This is where we must use base equalization technique. Let’s see how by doing our first problem on fractions.

Problem example 1

The rational number between $\displaystyle\frac{1}{2}$ and $\displaystyle\frac{3}{5}$ is,

  1. $\displaystyle\frac{2}{5}$
  2. $\displaystyle\frac{2}{3}$
  3. $\displaystyle\frac{4}{7}$
  4. $\displaystyle\frac{1}{3}$

Looking at the choices you get the main idea about what you have to do. The second given term being larger than the first, you need to select a number $x$ among the four choices such that,

$\displaystyle\frac{1}{2}<{x}<{\frac{3}{5}}$.

The longest solution

We may compare all these six (two given values and four choice values) fractions and form an ascending sequence. That would give us the desired answer. But as you can sense, it is possibly a crude and inefficient way of solving this problem. It would take much longer time and effort at the first glance.

Rather, you decide to test each option depending on its sensed relative value (near to $\frac{1}{2}$ or $\frac{3}{5}$) with either $\frac{1}{2}$ or $\frac{3}{5}$. The strategy that we would use here follows.

Strategy used

If the choice value being tested is smaller than $\frac{1}{2}$ or larger than $\frac{3}{5}$, we can straightway identify it to be outside the range from $\frac{1}{2}$ and $\frac{3}{5}$ and remove it from further consideration.

For example, identifying $\frac{1}{3}$ as a small number we compare it first with $\frac{1}{2}$, the lower value of the range. As $\frac{1}{3}$ is smaller than $\frac{1}{2}$, being out of range, it is eliminated from further consideration. We have used the first rule of fraction comparison in this case.

Fraction comparison rule 1

In a comparison between two fractions if both the numerators have same value, the fraction with lower denominator is larger (as the smaller denominator divides the same valued numerators).

Next we test $\frac{2}{5}$ as it seems to be small. Again we decide to compare it with $\frac{1}{2}$. But to be able to do it, we equalize the numerators by multiplying $\frac{1}{2}$ by $\frac{2}{2}$. It is transformed to $\frac{2}{4}$ with numerator equal to that of $\frac{2}{5}$. Knowing fraction rule 1 we reject $\frac{2}{5}$, it being out of range.

Aside:

Point 1: So we have used a technique of equalizing the numerator rather than the denominator, where numerator can be perceived as the Base of a fraction. Did we do something new? If you critically look at the activity, you would realize, the action of equalization is same - that is the abstract core activity. Depending on the situation, we conveniently decide to equalize the numerator or the denominator.

Point 2: Method of equalization: We simply multiplied the numerator and the denominator by the same value 2 as the method for transforming the numerator to its equal value 2. In general though method of equalization of numerator or denominator is to form target equal value as the LCM of the two and transform both the fractions. We would shortly see how.

Back to our problem we take up the next choice value $\frac{2}{3}$. By our previous experience we decide it to be fairly large. So we compare it with $\frac{3}{5}$ instead of $\frac{1}{2}$.

The numerators and denominators both are different and cannot easily be transformed to equal value by earlier simple method. Most cases of fraction manipulation are similar. In such cases invariably we resort to Base equalization technique with base as denominator. This is the standard technique for fraction manipulation.

Base equalization technique with fraction denominator as base

Let's take up a more general problem to clarify this concept. The problem we would solve is,

Problem 1.1. Evaluate, $\frac{3}{4} - \frac{5}{6} + \frac{7}{12}$.

Decision 1. First we would decide to equalize the denominators of all three fractions. We would shortly see why this technique works.

Decision 2. Method of equalization of the denominators is first to find the LCM of the denominators of all the fractions. That is the target equal value.

Action 1. Finding LCM: We find LCM of 4, 6 and 12 as 12 itself.

Action 2. Equalizing base denominator: To transform all three denominators to the single value of 12, their LCM, we multiply a fraction by $\frac{x}{x}$ where $x=\frac{LCM}{y}$, where $y$ is the denominator of the fraction being transformed.

To be specific we multiply $\frac{3}{4}$ by $\frac{3}{3}$, $\frac{5}{6}$ by $\frac{2}{2}$ and $\frac{7}{12}$ by $\frac{1}{1}$, getting respectively the transformed fractions, $\frac{9}{12}$, $\frac{10}{12}$ and $\frac{7}{12}$. The expression then will be,

$E = \frac{3}{4} - \frac{5}{6} + \frac{7}{12} = \frac{9}{12} - \frac{10}{12} + \frac{7}{12}$

Or, $E = \frac{1}{12} (9 - 10 + 7)$.

Reason behind the base equalization technique - why it works

We could factor out the problem creator (or barrier to the problem), the denominator as $\frac{1}{12}$ only because we have made all the denominators equal. As long as the denominators stay different we cannot do any further simple arithmetic operation. They act as the barriers to solving the problem. By using base equalization technique we remove this barrier outside the parenthesis and joyfully carry out the elementary arithmetic operations inside the parenthesis. Thus,

$E = \frac{1}{12} (9 - 10 + 7) = \frac{6}{12} = \frac{1}{2}$

Fraction comparison rule 2

In a comparison between two fractions if both the denominators have same value, the fraction with larger numerator will be larger (as the same valued denominator divides the larger numerator).

Returning back to our original problem now we are left with only one option value $\frac{4}{7}$. We know by reasoning that this will satisfy the given conditions, but we would be sure only when we do the tests. In this case, we have to test $\frac{4}{7}$, $\frac{1}{2}$ and $\frac{3}{5}$ to determine their relative values.

Following base equalization technique with base as denominator, we find the LCM of 2, 5 and 7 as 70 and transform the fractions as, $\frac{1}{2}$ to $\frac{35}{70}$, $\frac{4}{7}$ to $\frac{40}{70}$ and $\frac{3}{5}$ to $\frac{42}{70}$.

Clearly, according to the fraction comparison rule 2, the fraction $\frac{4}{7}$ of option c satisfies the given conditions. Option c is then the answer.

We will now formally state the preferred technique of base equalization for manipulation of fractions.

Preferred Base equalization technique for fraction manipulation

For comparison, addition and subtraction of fractions, transform the fractions so that the base denominator of each becomes the LCM of the denominators. Now carry out the required operations on these transformed fractions with ease after factoring out the inverse of common denominator.

We have already seen how this basic indispensable technique has been applied in all three elementary mathematical operations addition, subtraction and comparison on fractions. 

Reviewing what we have done till now we realize that we have considred numerical fractions only. But the concepts we have used are applicable for any general form of fractions with great positive results as well. We would take up an apparently complex problem to illustrate the general applicability of base equalization technique in fraction manipulation. From now on, we would generally assume in case of fraction manipulation - base means the denominator.

Problem example 2.

The value of,
$\frac{1}{a^2 +ax + x^2} - \frac{1}{a^2 - ax + x^2} +\frac{2ax}{a^4 + a^2x^2 + x^4}$

is,

  1. 2
  2. 1
  3. -1
  4. 0

Solution

Combining the first two terms

When you meet such balanced nearly whole square expressions in the denominator, one middle term of plus sign and the other minus, you can straightway pair the first and third terms $a^2$ and $x^2$ together so that when you combine the first two terms of the expression, the product of the denominators turns to $(a^2 + x^2)^2 - a^2x^2$.

We also knew from the nature of the denominators of the first two terms, that on subtraction in the numerator, the positive terms will be eliminated  and $-2ax$ will only be left as the resultant numerator. A practiced mind can do it fairly quickly.

Combining the result with third term

The result of this operation is,

$E = \frac{1}{a^2 +ax + x^2} - \frac{1}{a^2 - ax + x^2} +\frac{2ax}{a^4 + a^2x^2 + x^4}$

   $= \frac{(a^2 - ax + x^2) - (a^2 +ax + x^2)}{(a^2 + x^2)^2 - a^2x^2} + \frac{2ax}{a^4 + a^2x^2 + x^4}$

   $= \frac{-2ax}{a^4 + 2a^2x^2 + x^4 - a^2x^2} + \frac{2ax}{a^4 + a^2x^2 + x^4}$

   $= \frac{-2ax}{a^4 + a^2x^2 + x^4} + \frac{2ax}{a^4 + a^2x^2 + x^4} = 0$

Answer: Option d: 0.

Key concept used: Identifying regularities in the denominator first and then the numerator -- use of basic formula $(a + x)\times(a - x) = a^2 - x^2$ and $(a + x)^2$ formula to combine two terms and simplify  -- finally equalizing the denominators.

In this type of problems, key lies more often than not in identifying usable regular common patterns in the denominators. Once denominator complexity is resolved, numerator complexity automatically gets resolved.

As far as we are concerned, looking at the simple choice values, we were certain that finally the denominators can be simplified to an equal value, otherwise the overall complexity cannot be resolved to such a simple outcome as in the choice values. Simple algebraic manipulations produced the results, though underlying each term combining action, base equalization technique worked.

Abstraction

Before leaving now, we would note that in our previous use of Base equalization technique on indices problems, the actual method of equalization was different from its use in fraction problems, even though the activity in essence are the same, that is, equalization of the base. The methods of doing differed. This is where abstraction steps in.