The Plot Thickens

Doing maths looking at equations all day can sometimes get boring. Maths gets a lot more interesting when there are pictures. Most pictures in math involve graphs, so let’s start simple and begin with plotting points on a graph.

You are familiar with the number line:

You already know how to plot a point on this. But this is only a 1-dimensional plot. The most interesting thing you can plot on this is a horizontal line which represent all the numbers between the endpoints of the line. Wouldn’t it be nice if we could plot curves in 2-dimensions! Enter the cartesion coordinate system.

To the number line, let’s add a vertical number line intersecting the horizontal one at 0:

Each of these number lines are called an axis. The horizontal one is called the x-axis and the vertical one is called the y-axis. Now you can plot a point on any of these axes, but the strength of this system is that you can plot points anywhere on the surface that the coordinate system is on. But to plot a point in 2-dimensions, you need 2 numbers.

The typical way to indicate a point in maths is by using brackets. For example: (2, 1), (0, -7), (-4,3), (-1.5, -2.75). By convention, the first number is the x coordinate and the second number is the y coordinate, so the general point is (x, y). To plot a point, say (2, 1), you first go along the x axis 2 units to the right since 2 is positive, then go up 1 unit. That’s where the point (2, 1) is. I’ve plotted several other points below:

The point (0, 0) where the axes meet is called the origin. Note that positive x values are to the right of the origin and positive y values are above the origin. Negative values are to the left and below respectively.

Now this is cool but gets quickly boring just plotting points. The interesting things happen when we plot a set of points that satisfy an equation. I’ll get into that in my next post.

Algebra, Subscripts

In my tutoring travels, I notice that some students get confused when they see subscripts, for example \[{x}_{1}\]. As you know, there are only 26 letters in the alphabet. This is almost always enough to represent variables in algebra, but if a formula indicates a pattern, then this is difficult to do using just letters.A subscript is just a way of showing different unknowns using the same letter. \[{x}_{1}\] is a different unknown than \[{x}_{2}\] but the same letter x is used – only the subscript has changed. The subscript number just indicates an order and is not used in calculations.So for example, the method for finding the average of a set of numbers is to add up all the numbers, then divide by the number of numbers you just added. To show this in a maths formula:Average = \[\frac{{x}_{1}\hspace{0.33em}{+}\hspace{0.33em}{x}_{2}\hspace{0.33em}{+}\hspace{0.33em}{x}_{3}\hspace{0.33em}{+}\hspace{0.33em}\cdots\hspace{0.33em}{+}\hspace{0.33em}{x}_{n}}{n}\]Here, the pattern is easy to see. Each number in the set of numbers is given a different subscript. Since the subscript starts at 1 and ends in n, you can immediately see that there are n numbers, which is why the formula shows us dividing by n. The symbol “⋯” is called an ellipsis and indicates that you just follow the indicated pattern until you get to the last number, \[{x}_{n}\]. For any specific set of numbers, you know what n is, but since the formula is to apply for any set of numbers, we need to use the unknown n.Sometimes, the subscript is called an index. So in more complex formulas, you may see  \[{x}_{i}\] to represent any of the unknowns. So we could say that the average is the sum of all  \[{x}_{i}\]‘s divided by n.

 

Algebra, Word Problems

Nothing strikes fear in a math student like the dreaded word problem. I have seen many students who are very good at solving equations but do poorly with word problems. The problem is that they lack the skill to convert english into an equivalent math language. In my last post, I started with converting english phrases into algebraic expressions. Let’s graduate to a full word problem and create the equivalent algebraic equation.

Karen is twice as old as Lori. Three years from now, the sum of their ages will be 42. How old is Karen and Lori?

As I suggested in my last post, let’s break this down. So here we have two unknowns: Karen’s and Lori’s ages. So a good first step is to assign letters to these unknowns.  Let’s let K be Karen’s age and L be Lori’s age. Now the first sentence in the problem has a word that means “=” in math. That word is “is”.  In other word problems, you may see words like “the same as”, “equals”, “was”, “will be”.

The first sentence in the word problem directly converts to an equation since we already assigned letters to the two ages:

Karen is twice as old as Lori: K = 2L

Now there are two unknowns here but that’s OK. We can’t solve anything yet, but there is more information in the word problem. As we read it, write down the equivalent math expressions.

Three years from now, what are their ages three years from now? Well three years from now, Karen will be K + 3 and Lori will be L + 3.

the sum of their ages will be 42. Another equation here because of the words “will be”. So we add the ages of Lori and Karen three years from now to get 42:

(K + 3) + (L + 3) = 42

The brackets are not really needed. I just put them there so you can see that I am adding Karen’s age 3 years from now to Lori’s age 3 years from now.

Now we have the equation but there are two unknowns. You usually cannot solve a single equation  with more than one unknown. But remember the first equation we wrote down: K = 2L? This equation means that algebraically, K is exactly the same as 2L. In the second equation, we can replace the K with 2L:

(K + 3) + (L + 3) = 42: (2+ 3) + (L + 3) = 42

Now we can solve this equation to find what is. I covered solving equations before, so I won’t do a lot of explaining here. I will start by removing the brackets and proceed:

2+ 3 + L + 3 = 42

3L + 6 = 42

3L = 42 – 6 = 36

L = 36/3 = 12

So Lori is 12. What about Karen? Again, look at the things we’ve written down so far. We have K = 2L, that is, Karen is twice as old as Lori. Since we already know Lori’s age, Karen must be 24.

So in most word problems, it will help if you first assign letters to the unknowns, then create expressions and/or equations from each part of the word problem. Have these all together and usually, the equation you need to solve will pop out.

Algebra, The Beginnings

So let’s leave statistics for a while and return to algebra. This post will be a bit more basic but it illustrates a skill needed when converting word problems to equations. First, a couple of definitions:

An Expression in algebra is basically anything you can write down in algebra without the “equal” symbol. You can think of an expression as either side of an equation. Examples are

\[{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{-}{1}{,}\hspace{0.33em}{2}{a}\hspace{0.33em}{+}\hspace{0.33em}{3}{b}{,}\hspace{0.33em}\frac{{x}\hspace{0.33em}{+}\hspace{0.33em}{1}}{{x}\hspace{0.33em}{-}\hspace{0.33em}{1}}{,}\hspace{0.33em}{5}\]

They can be complex with more than one unknown or as simple as a number.

An Equation in algebra is two expressions with an “equal” symbol between them. Examples are

\[\begin{array}{l}{{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{3}{x}\hspace{0.33em}{-}{1}\hspace{0.33em}{=}\hspace{0.33em}{2}{a}\hspace{0.33em}{+}\hspace{0.33em}{3}{b}}\\{\frac{{x}\hspace{0.33em}{+}\hspace{0.33em}{1}}{{x}\hspace{0.33em}{-}\hspace{0.33em}{1}}\hspace{0.33em}{=}\hspace{0.33em}{5}}\\{{y}\hspace{0.33em}{=}\hspace{0.33em}{7}}\end{array}\]

So let’s look at creating expressions. In what follows, I am going to write an english phrase and follow that with the equivalent algebraic expression.

Double a number: 2x

One more than a number: x + 1

Half of a number: x/2

Seven less than triple a number: 3x -7

Take 5 more than a number then double it: 2(x + 5)

Other letters can be used to show an unknown number, but x is mostly used. Sometimes more that one unknown is needed:

Cost of 3 pears that cost each: 3p

Cost of 7 apples that cost a each: 7a

Total cost of the above fruit: 3p + 7a

The individual price 3 people pay at a restaurant if they split the bill: C/3

The total number of pencils in a classroom if each girl has 3 pencils and each boy has 2: 3g + 2b

It is usually a good practice to break down a word problem and write down the expressions first before generating an equation.

Let’s look at creating equations in my next post.

Quadratic Equations, Part 3

So we are in the midst of solving quadratic equations of the form:

\[
{ax}^{2}\hspace{0.33em}{+}\hspace{0.33em}{bx}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where a,b, and c are some numbers. So far, we have been looking at equations like this that can be factored so that we can use the Null Factor Law. But most quadratics cannot be factored by hand and the solutions to the equations are frequently decimal numbers. So what to do? Well fortunately mathematicians have heard your concerns and way back in the year 628, an Indian mathematician Brahmagupta came up with a formula called the quadratic formula. This formula will solve any quadratic equation.

So given a quadratic equation, if you identify the three coefficients a,b, and c, the solution can be obtained by using the following formula:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{b}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}}}{2a}
\]

This is a very powerful equation. It is not difficult to prove but I’d rather you research that yourself and for now, just accept  my word for it truthfulness. Let’s use it in some examples.

\[
{2}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{+}\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Here a = 2, b = 5, and c = 3. Let’s put these numbers in the quadratic formula. Notice that there is a symbol ± in the formula. That means that there are generally two solutions: one using the + symbol and the other using the – symbol. So the solutions to this equation are:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{5}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{2}{)(}{3}{)}}}{2(2)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}{1}}{4}\hspace{0.33em}{=}\hspace{0.33em}{-}{1}{,}\hspace{0.33em}{-}{1}{.}{5}
\]

So there are two solutions to this quadratic equation. Let’s do another one:

\[
{3}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Here a = 3, b = -4, and c = -7. It’s important to include the signs in the a,b, and and include them in the quadratic formula.

So the solutions are:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{(}{-}{4}{)}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{(}{-}{4}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{3}{)(}{-}{7}{)}}}{2(3)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{4}\hspace{0.33em}\pm\hspace{0.33em}{10}}{6}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{33}{,}\hspace{0.33em}{-}{1}
\]

Now look at the \[
{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}
\] part of the formula. If this is a positive number, like in the last two examples, then you can take the square root and get two solutions: one by using the + symbol and the other by using the – symbol. If this is zero, then you only get one solution as adding or subtracting a zero doesn’t give two solutions, just one. And if the expression is negative, well there is no solution in the real world because you can’t take the square root of a negative number. However, scientists and engineers work in a complex world where you can take the square root of a negative number, and this has a physical meaning. Maybe some day I will talk about these special numbers.

Quadratic Factoring, Part 2

Last time I introduced the method to factor some quadratic equations so that you can use the Null Factor Law to solve the equation. The example used was

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

which when factored became

\[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where you can readily see the two solutions x = 7 and -5. Let’s do another example.

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{24}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

So we know the factors will look like \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\] and we need to find the a and b so that the factored form is equivalent to the left side of the equation. a and b must be numbers that multiply to equal 24 and add or subtract to equal +2, the coefficient in front of the x. 8 and 3 do not work as they do not add or subtract to equal 2. However, 6 and 4 look like contenders.

So the signs of 6 and 4 must be such that they add to equal +2 but multiply to equal -24. Looks like +6 and -4 work so

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}{24}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where the two solution can be found by equating each factor to 0 which gives the two solutions x = 4 and -6.

Some quadratic equations may not be factorable but there is another method to solve these. I will show this method in my next post.

Now the generic quadratic equation looks like

\[
{a}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{b}{x}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where ab, and c are specific numbers. I have been merciful so far in that I have only looked at equations where a = 1. If it is some other value, it’s a little more difficult to factor, but it can be done. However, I will use the quadratic formula to work with these in my next post.

Quadratic Factoring

In my last post, I showed how to solve

\[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Now this is really a quadratic equation in disguise. When I covered the Distributive Property, you saw how to distribute a factor within a set of brackets:

\[
{x}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}
\]

This can also be done in reverse by un-distributing the x to get the expression back into factored form so that you can take advantage of the Null Factor Law. What I didn’t cover, was how to take an expression like \[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}
\] and un-factor it. To do this, you can distribute each term in the first set of brackets with each term in the second set:

\[
\begin{array}{l}
{{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}{(}{x}{+}{5}{)}\hspace{0.33em}{-}\hspace{0.33em}{7}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{-}{7}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}}\\
{{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}}
\end{array}
\]

So this is now recognisable as a quadratic expression. However, if I originally had the equation

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

we could not use the Null Factor Law. So what can you do if given this equation? There is a way to solve this using something called the Quadratic Formula, but that will be covered later. Here I will show how to factor this equation back to the original form we started with so that we can use the Null Factor Law.

So the goal is to take \[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}
\] and get it in the form

(something)(something else). If you look at how we unfactorised this, you can see that I can start with

\[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\], where we need to find the a and the b so that the expressions are equivalent. I know this is the way to start as the two x‘s are needed to get \[
{x}^{2}
\] when they are multiplied together. Now to find the a and the b, including the sign of each, you need to look at all the possible factors of the known number in the quadratic, in this case -35, that add up to the coefficient of the middle term, in this case -2. Again, this is suggested if you look at how we expanded \[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}
\]. The last number in the expansion (-35) is generated by multiplying the -7 and the +5. These two numbers also multiply the x‘s which are eventually added together to get the middle term, in this case -2x.

So to find the a and the b in \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\], let’s look at the possible factors of 35 (ignoring the sign for the moment): 35 and 1 or 7 and 5. 35 and 1 do indeed multiply to equal 35 but their addition or subtraction together do not equal 2. That leaves 7 and 5 which do satisfy both requirements: 7 × 5 = 35, 7 – 5 = 2. Now all that remains is to determine the signs. Since their multiplication has to equal -35, one of the numbers needs to be negative. And since the coefficient of the middle term is negative, the large number 7 needs to be negative as well. So in this case a = -7 and b = + 5 so that the factorisation we are looking for is \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\]. Now we can use the Null Factor Law to solve the equation, as we had done previously.

Not all quadratics can be factored like this, but this is a good skill to develop with practise. I will do several more examples in my next post.

Null Factor Law

I’d like to return to algebra and discuss the Null Factor Law which is useful in solving certain equations:

If two or more factors multiplied together equal zero, then the solutions can be found be equating each factor separately to zero.

This makes sense if you think of two numbers multiplied together equal 0:

\[
{ab}\hspace{0.33em}{=}\hspace{0.33em}{0}
\] can only be true if either a is zero, b is zero, or both are zero. No other non-zero numbers multiplied together can equal zero.

This is true for any algebraic expressions multiplied together. For example:

\[
{(}{x}{-}{7}{)(}{x}{+}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Can only be true if \[
{(}{x}{-}{7}{)\hspace{0.33em}=}\hspace{0.33em}{0}
\] or if \[
{(}{x}{+}{5}{)\hspace{0.33em}=}\hspace{0.33em}{0}
\]

Without formal algebra, you can see the two solutions to this equation are then x = 7 or -5.

Now this one was easy, but sometimes you are given an equation that is not a factored one:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

At first glance, it looks like the null factor law doesn’t apply here. But I did a post on the distributive property. Please review that if needed, but notice that there is a common factor of x in each of the terms on the left side of the equation. I can un-distribute this x to get:

\[
{x}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Looks like the null factor law can be used as there are now two factors on the left side. So mentally setting each of these to zero, we get the two solutions x = 0 or -5.

I covered a similar example on my post about quadratic equations. You can click on the tags on the right or below (depending on the device you are viewing this on) to directly go to previous posts on the listed topics.

I will be covering more complex examples in my next post.

Fractional Exponents Examples

In my last post, I ended with

\[
{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}\hspace{0.33em}{=}\hspace{0.33em}{\left({\sqrt[n]{x}}\right)}^{m}
\]

If I rewrite this, replacing the radical symbols with their exponent equivalents, I get

\[
{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}^{m}{)}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}^{\frac{1}{n}}{)}^{m}
\]. Note that \[
{m}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{n}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{n}\hspace{0.33em}\times\hspace{0.33em}{m}\hspace{0.33em}{=}\hspace{0.33em}\frac{m}{n}
\].

This seems to imply that if you have a base raised to a power and that is again raised to a power, you can simplify this by just multiplying the exponents. This is, in fact, true. You can show this for yourself by expanding

\[
{(}{2}^{3}{)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{{3}\times{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{6}
\].

In general,

\[
{(}{x}^{m}{)}^{n}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{m}\times{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{mn}
\].

By the way, \[
{x}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}
\] no matter how small or large x is, except for x = 0. \[
{0}^{0}
\] is not defined and if you try to do it, like dividing by 0, the math police will be knocking on your door.

Two other exponent rules, which you can demonstrate for yourself with simple examples are:

\[
{(}{xy}{)}^{m}\hspace{0.33em}{=}\hspace{0.33em}{x}^{m}{y}^{m}{,}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\left({\frac{x}{y}}\right)}^{m}\hspace{0.33em}{=}\hspace{0.33em}\frac{{x}^{m}}{{y}^{m}}
\]

With these rules and the ones covered previously, let’s do some examples:

  1. \[
    {(}{x}^{2}{y}{)(}{x}^{3}{y}^{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{+}{3}}{y}^{{1}{+}{4}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{5}{y}^{5}
    \]
  2. \[
    \frac{{x}^{2}{y}^{5}}{{xy}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{-}{1}}{y}^{{5}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{xy}^{2}
    \]
  3. \[
    {\left({7{a}^{3}{b}^{{-}{1}}}\right)}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}
    \]
  4. \[
    {\left({{z}^{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{z}^{{5}\times{2}}\hspace{0.33em}{=}\hspace{0.33em}{z}^{10}
    \]
  5. \[
    {\left({\frac{2x}{3{y}^{2}}}\right)}^{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}^{3}{x}^{3}}{{3}^{3}{y}^{{2}\times{3}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{8{x}^{3}}{27{y}^{6}}
    \]  (This problem uses several rules at the same time)
  6. \[
    {\left({{-}{2}{x}^{2}{y}^{{-}{4}}}\right)}^{{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{2}\times{(}{-}{2}{)}}{y}^{{-}{4}\times{(}{-}{2}{)}}}\right)\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{-}{4}}{y}^{8}}\right)\hspace{0.33em}{=}\hspace{0.33em}\frac{{y}^{8}}{4{x}^{4}}
    \]

If you have questions regarding any of these, send me an email via the Contact page, leave a comment via the Comments page, or post a comment on my facebook page.

Exponents and Roots

Well you might suspect that there is a connection between the roots of a number and exponents given my last couple of posts, and you would be correct. Now in my post on roots, I limited myself to using numbers that are perfect roots. For example, \[
\sqrt[3]{8}\hspace{0.33em}{=}\hspace{0.33em}{2}
\]. But many roots do not have such a simple answer. You just have to represent it exactly as a root. For example, \[
\sqrt{2}
\] has no simple integer solution. In fact, its decimal equivalent cannot be written down exactly as it is a non-repeating decimal. The answer you get on a calculator is just the first few decimals. \[
\sqrt{2}\] is an example of something called an irrational number. That doesn’t mean you can’t reason with it. It just means you can’t write it down exactly using decimals. You can only exactly represent it as \[
\sqrt{2}
\].

But it is still true that \[
\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}
\] (ignoring the negative solution for now).

Now from my post on exponents, you saw that

\[
{2}^{3}\hspace{0.33em}\times\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{{3}{+}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{5}
\]

Now bear with me, but consider

\[
{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{1}{2}{+}\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{1}\hspace{0.33em}{=}\hspace{0.33em}{2}
\].

But remember that

\[
\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}
\].

Perhaps

\[
{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{2}
\].

This is in fact true and the rules of exponents apply here as well. In general,

\[
{x}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{x}
\].

Now consider

\[
\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{3}{2}}
\].

Looks like a number raised to a fraction does have meaning. Now in decimals, this number is 2.82842712475… , which actually goes on forever. But on a calculator, you can get this answer by first raising 2 to the “3” power then taking the square root of the result, or first take the square root of 2 then raise the result to the “3” power. In general:

\[
{x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}\hspace{0.33em}{=}\hspace{0.33em}{\left({\sqrt[n]{x}}\right)}^{m}
\].

In my next post, I will show several examples of fractional exponents.