# Chaitanya's Random Pages

## April 22, 2011

### Manipulating Fractions

Filed under: mathematics — ckrao @ 7:40 am

This post is partly inspired by a fractions quiz I recently tried on Sporcle. It made me think of the ways I know of manipulating fractions (whether they be rational numbers or algebraic expressions), so I decided to collect them here.

1. General form

$\displaystyle \frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$

In particular $\displaystyle \frac{1}{a} \pm \frac{1}{b} = \frac{b \pm a}{ab}$, $\displaystyle \frac{1}{a} + \frac{1}{a} = \frac{2}{a} = \frac{1}{a/2}$ if a is even.

2. Find a small common multiple of b and d: let it be bd if you can’t think of anything smaller.

e.g. $\displaystyle \frac{3}{4} + \frac{1}{6} = \frac{3.3 + 1.2}{12} = \frac{11}{12}$

3. Equivalently, take out any common factor you see in the denominator.

$\displaystyle \frac{3}{4} + \frac{1}{6} = \frac{1}{2}\left(\frac{3}{2} + \frac{1}{3}\right) = \frac{1}{2}.\frac{11}{6} = \frac{11}{12}$

4. Of course you can also take out any common factor in the numerator:

e.g. $\displaystyle \frac{4}{8} + \frac{4}{7} = 4\left(\frac{1}{8} + \frac{1}{7}\right) = 4 \times \frac{15}{56} = \frac{15}{14}$

(Moral: the distributive law is your friend!)

5. It may help to simplify the fraction first:

e.g. $\displaystyle \frac{1}{4} + \frac{5}{10} =\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

If you are good with decimals or percentages, in some cases it’s easier to convert to that domain:

e.g. $\displaystyle \frac{2}{5} + \frac{1}{4} = 40\% + 25\% = 65\% = \frac{13}{20}$

Mixed fractions: separate the integer and fractional parts:

e.g. $\displaystyle 1 \frac{1}{2} - 3 \frac{3}{4} = \left(1 - 3\right) + \left(\frac{1}{2} - \frac{3}{4}\right) = -2 - \frac{1}{4} = -2\frac{1}{4} \left( = -\frac{9}{4}\right)$

Multiplication and division

6. Get rid of mixed fractions first! e.g. $\displaystyle 2 \frac{2}{3} = \frac{8}{3}$

7. General form

$\displaystyle \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$

$\displaystyle \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$

8. Find common factors to allow for cancellation in the numerator and denominator

$\displaystyle \frac{ka}{kb} = \frac{a}{b}$

9. You can always multiply or divide both numerator and denominator by the same quantity.

e.g. $\displaystyle \frac{a/b}{c/b} = \frac{a}{c}$

Application 1: Rationalising the denominator.

e.g. $\displaystyle 1/\sqrt{2} = \sqrt{2}/2$

e.g. $\displaystyle \frac{1}{2-\sqrt{3}} = \frac{2 + \sqrt{3}}{\left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right)} = \frac{2 + \sqrt{3}}{4-3} = 2 + \sqrt{3}$

Application 2: Manipulating factorials to work with combinatorial expressions.

e.g. $\displaystyle 10 \times 9 \times 8 \times 7 = \frac{10!}{6!} = 4! \binom{10}{4}$

Application 3: Manipulation of algebraic expressions

e.g. $\displaystyle \frac{\frac{1}{x} - \frac{1}{y}}{\frac{1}{x} + \frac{1}{y}} = \frac{y-x}{y+x}$

(Here we multiplied the numerator and denominator by xy to perform the above in one step.)

Equations

10. If you have an equation with x in the denominator, it can be swapped with the right side in one step.

e.g. $\displaystyle \frac{3}{x} = 4 \Rightarrow\frac{3}{4} = x$

(Note how the x and 4 are simply swapped by multiplying both sides by x/4.)

Ratios

Something I did not learn in regular school classes is that if a/b = c/d, then both expressions are also equal to (a+c)/(b + d). This manipulation is known as addendo, and there are several other terms when working with ratios. I found the following terminology mostly in [1].

• If I write the ratio a:b or a/b or$\frac{a}{b}$, the a is known as the antecedent, and b (non-zero) the consequent.
• The slash is called a solidus and the horizontal fraction bar the vinculum.
• The duplicate ratio of a:b is$a^2:b^2$ and similarly the triplicate ratio is$a^3:b^3$.
• The sub-duplicate and sub-triplicate ratios are given by $a^{1/2}:b^{1/2}$ and $a^{1/3}:b^{1/3}$ respectively.
• A ratio is commensurate if it is rational.
• A continued ratio joins 3 or more quantities, e.g. 1:2:3.
• An equality of two ratios is called proportion$a:b = c:d \Rightarrow ad = bc$. (This is called the cross product rule.) The a and d are called extremes and the b and c are called means (hence product of extremes equals product of means)
• If a:b = b:c, b is called the mean proportional between a and c, and$b = \pm\sqrt{ac}$. In this expression, a is called the first proportional and c the third proportional.

Invertendo: If a:b = c:d then b:a = d:c.

Alternendo: If a:b = c:d then a:c = b:d.

Componendo: If a:b = c:d then (a+b):b = (c+d):d.

Dividendo: If a:b = c:d then (a-b):b = (c-d):d.

Componendo and dividendo: If a:b = c:d then (a+b):(a-b) = (c+d):(c-d).

Addendo: If a:b = c:d = e:f = … then each of these is equal to (a+c+e + …):(b+d+f+ …).

Subtrahendo: If a:b = c:d = e:f = … then each of these is equal to (a-c-e-…):(b-d-f-…).

Crescendo: An increasing sequence of ratios.

Innuendo: A subtle hint of a ratio.

Nintendo: A ratio you like playing with.

The last three were jokes by the way. 🙂

I wonder how many of the above terms I will use in future!

11. Back to addendo, more generally the ratios can be weighted, so if$a_1:b_1 = a_2:b_2 = \ldots = a_n:b_n$ then each of these is equal to

$\displaystyle \frac{\sum_{i=1}^n \lambda_i a_i}{\sum_{i=1}^n \lambda_i b_i}$.

A nice application of addendo/subtrahendo is in the following proof of Ceva’s theorem: let ABC be a triangle with P a point inside. Extending AP, BP and CP to meet the triangle’s sides at D, E and F as in the figure, we have the relationship

$\displaystyle \frac{AF}{FB}.\frac{BD}{DC}.\frac{CE}{EA} = 1.$

Proof: denoting the area of triangle XYZ as |XYZ|, we write the ratio of sides in terms of a ratio of areas of two sets of triangles, then apply subtrahendo:

$\displaystyle \frac{AF}{FB} = \frac{|ACF|}{|BCF|} = \frac{|APF|}{|BPF|} = \frac{|ACF|-|APF|}{|BCF|-|BPF|} = \frac{|ACP|}{|BCP|}.$

Similarly,$\displaystyle \frac{BD}{DC} = \frac{|ABP|}{|ACP|}$ and $\displaystyle \frac{CE}{EA} = \frac{|BCP|}{|ABP|}$. Multiplying these three ratios together gives the desired result.

A second application is in solving this question I found recently on math.stackexchange.com:

If$\displaystyle M(x_2,y_2)$ is the foot of a perpendicular drawn from$\displaystyle P(x_1,y_1)$ on the line$ax + by + c = 0$ (a and b non-zero), then we have

$\displaystyle \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{-\left(ax_1 + by_1 + c\right)}{a^2 + b^2}.$

The first two equality follows from MP being perpendicular to the line having slope -b/a. To obtain the second equality we use addendo:

$\displaystyle \frac{x_2 - x_1}{a} = \frac{a(x_2 - x_1)}{a^2} = \frac{b(y_2 - y_1)}{b^2} = \frac{a(x_2 - x_1) + b(y_2 - y_1)}{a^2+b^2} = \frac{-\left(ax_1 + by_1 + c\right)}{a^2 + b^2},$

where the last equality follows from the fact that $\displaystyle M(x_2,y_2)$ is on the line and hence satisfies $ax_2 + by_2 + c = 0$. Nice!

Linear Fractional Transformations

12. If you have an equation like $\displaystyle y = \frac{1-x}{1+x}$ and you want to write x in terms of y, you can perform a series of manipulations to get there in maybe four or five steps, or use the following fact:

The inverse of $\displaystyle y = \frac{ax + b}{cx +d}$ is $\displaystyle x = \frac{dy - b}{-cy + a}$.

This is easy to remember if you know that the inverse of the 2 by 2 matrix $\displaystyle \left[\begin{array}{cc}a&b\\c&d\end{array}\right].$ is a multiple of $\displaystyle \left[\begin{array}{cc}d&-b\\-c&a\end{array}\right]$ (swap the diagonals, change the sign of the off-diagonals). Functions of the above form are known as fractional linear transformations, and are very interesting objects to work with. There is in fact a strong connection between the function $\frac{ax + b}{cx +d}$ and the matrix $\displaystyle \left[\begin{array}{cc}a&b\\c&d\end{array}\right].$ It can be verified by direct computation that the composition of two fractional linear transformations corresponds to the product of the corresponding matrices.

Another way to see why this correspondence works is as follows (refer to [2, pp27-8] for more information). The real projective line $\displaystyle \mathbb{RP}^1$ can be viewed as a pair of real numbers $\left(\begin{array}{c}x_1\\x_2\end{array}\right)$, where $\left(\begin{array}{c}x_1\\x_2\end{array}\right)$ and $\left(\begin{array}{c}kx_1\\kx_2\end{array}\right)$ are equivalent. An element of $\displaystyle \mathbb{RP}^1$ can be a viewed as a 1-dimensional subspace in $\mathbb{R}^2$ with the origin removed. There is a bijective correspondence between $\displaystyle \mathbb{RP}^1$ and the real line plus infinity, given by

$\displaystyle \left(\begin{array}{c}x_1\\x_2\end{array}\right) \leftrightarrow \frac{x_1}{x_2}$ if $x_2 \neq 0, \quad \left(\begin{array}{c}x_1\\ 0 \end{array} \right) \leftrightarrow \infty$.

Any invertible linear transformation of the real plane $\mathbb{R}^2$ maps any line to a line. In particular it induces a (bijective) map from elements of $\displaystyle \mathbb{RP}^1$ to elements of $\displaystyle \mathbb{RP}^1$ (which we recall are viewed as 1-dimensional subspaces). Hence any invertible linear transformation of the real plane corresponds to a bijection from the real line plus infinity to itself.

If M is such a transformation, it can be represented by left multiplication by a 2 by 2 matrix $\displaystyle \left[\begin{array}{cc}a&b\\c&d\end{array}\right]$. Let $\phi$ be the bijection of the extended real line induced by M. If $x \in \mathbb{R}$, then M sends  $\displaystyle \left[\begin{array}{c}x\\1 \end{array}\right]$ to $\displaystyle \left[\begin{array}{c}ax+b\\cx+d \end{array}\right]$, which implies $\phi(x) = \frac{ax+b}{cx+d}$.

Hence the linear fractional transformations have a group structure and under compositions may be represented by matrix multiplications. This is why their inverse can be determined easily.

In particular, if a = -d (i.e. coefficient of x in numerator and constant coefficient in denominator sum to zero), then

$\displaystyle f(x) = \frac{ax+b}{cx-a}$ is its own inverse.

For example, $\displaystyle y = \frac{1-x}{1+x} \Rightarrow x = \frac{1-y}{1+y}$.

Partial Fractions

This is the about the techniques to reduce a ratio of polynomials into a sum of simpler ratios. It comes up most often in integration and complex analysis. If the degree of the numerator is at least that of the denominator we can always use division to have the remaining numerator of degree less than that of the denominator. We assume this to be the case from now.

The theory is large, so here I will mention only the most common occurrences:

(A) Linear factors in the denominator: $\displaystyle \frac{f(x)}{\prod_{i=1}^n(x-a_1)^{\alpha_i}} = \sum_{i=1}^n \sum_{j = 1}^{\alpha_i} \frac{A_{ij}}{(x-a_i)^j}$.

The question: how do we find $\displaystyle A_{ij}$?

(B) The denominator has quadratic factors which cannot be factored over the reals: $\displaystyle \frac{f(x)}{\prod_{i=1}^m(x-a_i)^{\alpha_i}\prod_{i=1}^n(x^2-b_ix +c_i)^{\beta_i}} = \sum_{i=1}^m \sum_{j = 1}^{\alpha_i} \frac{A_{ij}}{(x-a_i)^j} + \sum_{i=1}^m \sum_{j = 1}^{\beta_i} \frac{B_{ij}x + C_{ij}}{(x^2-b_ix +c_i)^j}$.

13. For (A), the easiest case is when all the $\alpha_i$ values are 1. We simply multiply both sides by $(x-a_i)$ and then set $x=a_i$ to isolate $A_{i1}$.

e.g. $\begin{array}{lcl} \frac{1}{x(x-1)(x-2)} &=& \frac{1/(-1.-2)}{x} + \frac{1/(1.-1)}{x-1} + \frac{1/(2.1)}{x-2}\\ &=& \frac{1}{2x} - \frac{1}{x-1} + \frac{1}{2(x-2)}\end{array}$

(This can be done straight away, more or less.)

14. If one of the factors on the denominator has $\alpha_i > 1$, I work out the other numerators first, then bring them to the other side to find the final coefficient.

For example, to split $\frac{1}{x(x-1)^2}$ we can initially write

$\displaystyle \frac{1}{x(x-1)^2} = \frac{1/(0-1)^2}{x} + \frac{1/1}{(x-1)^2} + \frac{A}{x-1}.$

This becomes

$\displaystyle \frac{A}{x-1} = \frac{1 - (x-1)^2 - x}{x(x-1)^2} = \frac{(x-1)(1-x-1)}{x(x-1)^2} = \frac{-1}{x-1},$

from which A = -1 and we conclude

$\displaystyle \frac{1}{x(x-1)^2} = \frac{1}{x} - \frac{1}{x-1}+ \frac{1}{(x-1)^2}$.

15. In the more general case of (A), we multiply both sides by $(x-a_i)^j$, move terms with powers of $(x-a_i)$ in the denominator to the left side, and then take the limit of both sides as $x \rightarrow a_i$. This limit will require $\alpha_i-i$ derivatives (applications of l’Hôpital’s rule) as both numerator and denominator of the left side have the root $a_i$ with multiplicity $\alpha_i-i$. The result is the following form.

$\displaystyle A_{ij} = \frac{1}{(\alpha_i-i)!} \frac{d^{\alpha_i-i}}{dx^{\alpha_i-i}} \frac{f(x)(x-a_i)^{\alpha_i}}{\prod_{k=1}^n(x-a_1)^{\alpha_i}}$.

16. Finally, to deal with case (B), one could either use complex numbers as roots of the quadratics and proceed as in the linear case, or multiply through by the common denominator, and equate the coefficients of like terms. This forms a system of linear equations which can be solved (this general method is probably how computer packages tackle the problem).

17. Once again, if there is only one “nuisance factor”, it is easiest to take other terms to the left.

e.g. to split $\frac{1}{x^3 + 1}$, we first write

$\displaystyle \frac{1}{x^3 + 1} = \frac{1/((-1)^2 - (-1) + 1)}{x+1} + \frac{Ax + B}{x^2 - x + 1}.$

Then $\displaystyle \frac{Ax + B}{x^2 - x + 1} = \frac{3 - (x^2 - x + 1)}{3(x+1)(x^2 - x+ 1)} = \frac{(x+1)(2-x)}{3(x+1)(x^2 - x+ 1)} = \frac{2-x}{3(x^2 - x+ 1)}$. Hence A = -1/3, B = 2/3, and

$\displaystyle \frac{1}{x^3 + 1} = \frac{1}{3(x+1)} + \frac{2-x}{3(x^2 - x+ 1)}$.

#### References

[1] N.V. Ravi, Ratio and Proportion, http://www.icai.org/resource_file/16808Ratio-Proportion.pdf

[2] D. Sarason, Complex function theory, AMS (2nd edition), 2007.