Unit About Units

Besides coming up with the equations to solve in word problems, another task is often to convert units of measurement. For example, what is the area of a rectangle 3 feet long and 10 centimetres wide. The area of a rectangle is the two dimensions multiplied together. The problem here is that one dimension is in feet and the other in centimetres. The units of area for this problem should be either feet-squared (ft²) or centimetres-squared (cm²).

So you look up the conversion between feet and centimetres and see that there are 30.48 centimetres in a foot. What do I do with this number?

So if the area is to be expressed in cm², you need to convert the 3 feet to centimetres. Do I multiply or divide the 3 feet by 30.48? One way to determine this is to think of the relative sizes of the units. A centimetre is smaller than a foot, so you would expect a dimension in feet to be a bigger number when expressed in centimetres. That tells you to multiply and you would get 3 ft × 30.48 = 91.44 cm and that is correct.

Conversely, if you wanted to convert the centimetres to feet, since feet are larger than centimetres, you would divide the centimetres dimension: 10 cm ÷ 30.48 = 0.33 ft (rounded to 2 decimal places).

Another way to determine what to do is to look at the units themselves. Units in numbers behave just like variables in that they can be added, subtracted, multiplied, divided, and cancelled using the same rules as with variables. For example, if I walked 2 km, stopped, then walked 500 m, how far did I walk? Well you just can’t add the numbers together as 502 would be in what units? 2 km cannot be directly added to 500 m just as 2x + 500y can’t be added because the variables are different. But if I convert one of the units to the other so that the units are the same, then I can add the numbers. 500 m is 0.5 km so: 2 km + 0.5 km = 2.5 km. This works for the same reason that 2x + 0.5x = 2.5x works.

Now let’s look at the ft/cm conversions by looking at the units. The conversion number 30.48 has units as well. There are 30.48 cm per ft or 30.48 cm/ft. The units part can be seen as a fraction with cm in the numerator and ft in the denominator. So if we want to convert ft to cm, we want to form an expression so that the “ft” part cancels just like a variable would:

\[{3}\hspace{0.33em}\rlap{-}{\mathrm{ft}}\hspace{0.33em}\times\hspace{0.33em}{30}{.}{48}\frac{\mathrm{cm}}{\rlap{-}{\mathrm{ft}}}\hspace{0.33em}{=}\hspace{0.33em}{91}{.}{48}\hspace{0.33em}{\mathrm{cm}}\]

If we want to convert the cm to ft, then we would want to cancel the cm part. The conversion number can also be interpreted as being 1 ft per 30.48 cm or 1 ft/30.48 cm. Multiplying by this fraction, effectively divides the 10 cm by 30.48 and the “cm” units cancel:

\[{10}\hspace{0.33em}\rlap{-}{\mathrm{c}}\rlap{-}{\mathrm{m}}\hspace{0.33em}\times\hspace{0.33em}\frac{{1}\hspace{0.33em}{\mathrm{ft}}}{{30}{.}{48}\hspace{0.33em}\rlap{-}{\mathrm{c}}\rlap{-}{\mathrm{m}}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{33}\hspace{0.33em}{\mathrm{ft}}\]

So now, the area of the rectangle can be found:

91.44 cm × 10 cm = 914.4 cm × cm = 914.4 cm²

or

3 ft × 0.33 ft = 0.99 ft × ft = 0.99 ft²

This is why area units are “squared”.

The Mostly Long and Short About Division

Long division troubles some of my students, so here is a short tutorial on how to do it.

Let’s look at 356 ÷ 23. To set this up to do long division, you use the rather descriptive long division symbol,

The number you divide into (356) goes inside the symbol. In this form, it is called the dividend. The other number is called the divisor. To begin long division, starting from the left side of 356, what is the first number that 23 divides into? 35 is the first number and 23 goes into that 1 times. Put the 1 above the 5.

Now multiply the 1 times the 23, put the result under the 35 then subtract. You should get 12,

Now bring down the next digit so that 126 is now showing,

Now repeat the above steps with the 126. How many times does 23 go into 126. You can guess. If after you multiply, you get a number greater than 126, you guessed too high. If after you subtract, you get a number greater than 23, you guessed too low,

Too high,

Too low,

Just right. You are done as there are no more digits to drop down. The answer is 15 with a remainder of 11. If there is no remainder, then you would get a 0 at the bottom.

Now if you did this problem on a calculator, you would get 15.478… . The numbers to the right of the decimal point are the remainder expressed as a decimal number. You can get this answer as well using long division by adding a decimal point to the dividend and adding zeros to the right, as many as required for the accuracy needed. Then just continue the above process,

 

An Irrational Post

In my posts to date, you have seen different types of numbers: Positive numbers, Negative numbers, Integers, Fractions, and Prime numbers. Some numbers can fit into more than one category. Today I want to introduce two more types: Rational and Irrational numbers.

It is easier to define what a rational number is then anything that is not rational, is an irrational number.

A rational number is any number that can be put into the form

\[\frac{a}{b}\]

where a and b are integers. So all fractions are rational numbers, but so are all  the integers like 0, -1, 3, -9,999 and so on because they can all be represented as the integer over “1”:

\[\frac{0}{1}{,}\hspace{0.33em}\frac{{-}{1}}{1}{,}\hspace{0.33em}\frac{3}{1}{,}\hspace{0.33em}\frac{{-}{9}{,}{999}}{1}\]

Now decimal numbers can be rational as well. The way to tell is if the decimal part repeats a pattern no matter how short or long the pattern is. That is because all repeating pattern decimals can be put into the a/b form. So the following are all rational numbers. Note that a line over a section of the decimal means that that section is repeated over and over: 

\[{3}{.}{0}\overline{0}{,}\hspace{0.33em}{0}{.}{111}\overline{1}{,}\hspace{0.33em}{6}{.}{25}\overline{25}{,}\hspace{0.33em}{-}{2}{.}{1275}\overline{1275}\]

That is in fact another way to test for whether a number is rational: if the decimal part (even if it is “0”) repeats eventually, it is a rational number. Any number that cannot be expressed as a repeating decimal is irrational. You have been exposed to one famous irrational number: 𝜋. This number has a non-repeating decimal that goes on forever without repeating. This has been proven to be true many times. Other numbers that are irrational are many (but not all) square roots and other roots like cube roots. For example, the following are also irrational numbers:

\[\sqrt{2}{,}\hspace{0.33em}\sqrt{3}{,}\hspace{0.33em}\sqrt[3]{7}{,}\hspace{0.33em}\sqrt[4]{2}\]

In fact, using the square root symbols or other symbols like 𝜋 are the only way we can express the number exactly. Even modern calculators with many digits of accuracy can only represent an irrational number to a limited number of decimal places. So when using an irrational number in a calculation, you use the number of decimal places required for the accuracy required in the final answer.

The set of rational numbers and irrational numbers comprise all the numbers on the number line. This complete set of numbers on the number line are called real numbers.

There are many other types of numbers. For example transcendental numbers and things called imaginary numbers. So we have real numbers and imaginary numbers. These can be combined to form another type of number, complex numbers. These are used in many branches of engineering and have physical meaning, even though imaginary numbers are used. Imaginary numbers are not on the real number line, they are plotted using another different number line. Maybe I will cover these later when I have a real, rational moment.

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.

 

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.

Exponent Rules

So I will get back to roots of numbers, but let’s first look at the rules for combining exponents.

Now an integer exponent means that you are multiplying the base together as many times as the exponent indicates. That is,

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

The exponent “5” says to multiply the base “3” five times. This immediately suggests our first rule of exponents:

\[
{x}^{m}{x}^{n}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{m}{+}{n}}
\]

That is, when the same base with exponents are multiplied together, you can simplify this by adding the exponents. You can readily see this with the example above:

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

So this rule makes sense. Now let’s look at another example to motivate the next exponent rule:

\[
\frac{{3}^{3}}{{3}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{3}}{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{1}\hspace{0.33em}{=}\hspace{0.33em}{3}^{1}
\]

Now you would normally leave out the exponent “1” in the final answer but I left it there so you can see the following rule in action:

\[
\frac{{x}^{m}}{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{m}{-}{n}}
\]

Can you see how this rule works for the last example? By the way, these rules work whether or not the base is a known number or not.

\[
{x}^{13}{x}^{7}\hspace{0.33em}{=}\hspace{0.33em}{x}^{20}{,}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{{x}^{13}}{{x}^{7}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{6}
\]

 

Now so far, I have limited myself to using positive integers as the exponents. It turns out that any number can be used as an exponent but it is not clear what a negative or non-integer exponent means. Let’s first look at negative integer exponents.

In the division example above with the base “3”, I specifically put the “3” with the larger exponent in the numerator. What if I reversed these:

\[
\frac{{3}^{2}}{{3}^{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}}{\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{3}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{3}^{1}}
\]

But according to the division rule of exponents:

\[
\frac{{3}^{2}}{{3}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{3}^{{2}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{3}^{{-}{1}}
\]

This suggests that \[
{3}^{{-}{1}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{3}^{1}}
\] and this is correct. A negative exponent of a base means the equivalent to the same base to the positive exponent in the denominator. It actually works in the other direction as well. You can move factors between the numerator and denominator as long as you change the sign of the exponent:

\[
\begin{array}{l}
{{x}^{{-}{6}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{{x}^{6}}}\\
{\frac{{x}^{{-}{6}}\hspace{0.33em}{y}^{5}}{{z}^{{-}{7}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{y}^{5}{z}^{7}}{{x}^{6}}}
\end{array}
\]

And the multiplication rule works as well:

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

That takes care of the numbers on the tick marks of the number line as exponents, but what about the numbers in between? That will be the topic of my next post.

Back to our Roots

I am tired of fractions, you too? Well let’s switch gears and talk about the roots of numbers. This is preparing you for a post or posts on the rules of exponents.

So before, I introduced the concept of the square root and how it is the reverse operation of squaring a number:

\[
\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}\pm{5}
\] because \[
{5}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}
\]. That is \[
\sqrt{{5}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{5}
\]. Or in general, \[
\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\pm{x}
\]. Remember that when taking the square root, there are two solutions since \[
{\left({{-}{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}
\] as well.

Well, what about the opposite operation to \[
{x}^{3}
\]? Well there is one:

\[
\sqrt[3]{{x}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}
\]

Notice a few things here. First, there is only one solution. There are no plus/minus solutions because the index (the “3”) of the root is odd and using the rules of signs, the sign of the odd root of a number will be the same as the number in the radical symbol. That is:

\[
\sqrt[3]{125}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}\hspace{0.33em}\sqrt[3]{{-}{125}}\hspace{0.33em}{=}\hspace{0.33em}{-}{5}
\]

The other thing to notice is that if there is no index shown, then a “2” is assumed to be there. If the index is a “3”, it is called a cube root. After that, you use ordinal numbers, that is fourth root, fifth root, etc.

The last thing to notice is if the index is even, then you do get two plus/minus solutions. If the index is odd, you only get one solution.

So in general,

\[
\sqrt[n]{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}\pm{x}
\] if n is even and

\[
\sqrt[n]{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}
\] if n is odd.

Examples:

\[
\begin{array}{l}
{\sqrt[4]{16}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[4]{{(}\pm{2}{)}^{4}}\hspace{0.33em}{=}\hspace{0.33em}\pm{2}}\\
{\sqrt[3]{{-}{8}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[3]{{(}{-}{2}{)}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{-}{8}}\\
{\sqrt[5]{32}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[5]{{2}^{5}}\hspace{0.33em}{=}\hspace{0.33em}{2}}\\
{\sqrt[5]{{-}{32}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[5]{{(}{-}{2}{)}^{5}}\hspace{0.33em}{=}\hspace{0.33em}{-}{2}}\\
{\sqrt[4]{81}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[4]{{(}\pm{3}{)}^{4}}\hspace{0.33em}{=}\hspace{0.33em}\pm{3}}
\end{array}
\]

 

In my next post, I’ll introduce some rules regarding exponents.

Fractions Part 9 – Fraction Problems

OK, let’s now use the skills in my previous posts to solve some example problems. I will show the problems first so you can try them on your own, then further down the post, you can see the solutions:

\[
\begin{array}{l}
{{1}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\div\hspace{0.33em}{2}\frac{7}{8}}\\
{{2}{.}\hspace{0.33em}\hspace{0.33em}\frac{7}{9}\hspace{0.33em}{+}\hspace{0.33em}\frac{11}{12}}\\
{{3}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}{+}\hspace{0.33em}{2}\frac{7}{8}}\\
{{4}{.}\hspace{0.33em}\hspace{0.33em}\frac{11}{12}\hspace{0.33em}{-}\hspace{0.33em}\frac{7}{9}}\\
{{5}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\times\hspace{0.33em}{2}\frac{7}{8}}
\end{array}
\]

 

 

 

 

 

 

 

 

 

 

SOLUTIONS:

\[
{1}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\div\hspace{0.33em}{2}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}\div\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{8}{23}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{8}}{{4}\hspace{0.33em}\times\hspace{0.33em}{23}}
\]

 

Now I will simplify this as much as possible now before I multiply to keep the numbers as small as possible:

\[
\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{8}}{{4}\hspace{0.33em}\times\hspace{0.33em}{23}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}\rlap{/}{4}\hspace{0.33em}\times\hspace{0.33em}{2}}{\rlap{/}{4}\hspace{0.33em}\times\hspace{0.33em}{23}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{2}}{23}\hspace{0.33em}{=}\hspace{0.33em}\frac{50}{23}\hspace{0.33em}{=}\hspace{0.33em}{2}\frac{4}{23}
\]

 

\[
{2}{.}\hspace{0.33em}\hspace{0.33em}\frac{7}{9}\hspace{0.33em}{+}\hspace{0.33em}\frac{11}{12}
\].  Since these fractions are being added and they have different denominators, we need to find a common denominator. To find the Least Common Denominator (LCD):

9 = 3 × 3,   12 = 2 × 2 × 3

LCD = 2 × 2 × 3 × 3 = 36

So now convert each of the fractions into equivalent ones with 36 as the denominator. So I will multiply the top and bottom of the first fraction by 4 and the other one by 3:

\[
\frac{7}{9}\hspace{0.33em}{+}\hspace{0.33em}\frac{11}{12}\hspace{0.33em}{=}\hspace{0.33em}\frac{{7}\hspace{0.33em}\times\hspace{0.33em}{4}}{{9}\hspace{0.33em}\times\hspace{0.33em}{4}}\hspace{0.33em}{+}\hspace{0.33em}\frac{{11}\hspace{0.33em}\times\hspace{0.33em}{3}}{{12}\hspace{0.33em}\times\hspace{0.33em}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{28}{36}\hspace{0.33em}{+}\hspace{0.33em}\frac{33}{36}\hspace{0.33em}{=}\hspace{0.33em}\frac{61}{36}\hspace{0.33em}{=}\hspace{0.33em}{1}\frac{25}{36}
\]

 

\[
{3}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}{+}\hspace{0.33em}{2}\frac{7}{8}
\]. First, the fractions need to be converted to improper ones. Then since 4 evenly divides into 8, we just need to convert the first fraction to an equivalent one with 8 as the denominator:

\[
{6}\frac{1}{4}\hspace{0.33em}{+}\hspace{0.33em}{2}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}{+}\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{{25}\hspace{0.33em}\times\hspace{0.33em}{2}}{{4}\hspace{0.33em}\times\hspace{0.33em}{2}}\hspace{0.33em}{+}\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{50}{8}\hspace{0.33em}{+}\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{73}{8}\hspace{0.33em}{=}\hspace{0.33em}{9}\frac{1}{8}
\]

 

\[
{4}{.}\hspace{0.33em}\hspace{0.33em}\frac{11}{12}\hspace{0.33em}{-}\hspace{0.33em}\frac{7}{9}
\]. These are the same fractions as in the second problem, but now it’s a subtraction problem. We have already converted these two fractions into equivalent fractions with the same denominator, so we will use that result here:

\[
\frac{11}{12}\hspace{0.33em}{-}\hspace{0.33em}\frac{7}{9}\hspace{0.33em}{=}\hspace{0.33em}\frac{33}{36}\hspace{0.33em}{-}\hspace{0.33em}\frac{28}{36}\hspace{0.33em}{=}\hspace{0.33em}\frac{5}{36}
\]

 

\[
{5}{.}\hspace{0.33em}\hspace{0.33em}{6}\frac{1}{4}\hspace{0.33em}\times\hspace{0.33em}{2}\frac{7}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{25}{4}\hspace{0.33em}\times\hspace{0.33em}\frac{23}{8}\hspace{0.33em}{=}\hspace{0.33em}\frac{575}{8}\hspace{0.33em}{=}\hspace{0.33em}{71}\frac{7}{8}
\]

 

How did you do?

Fractions Part 8 – Fraction Division

So far I have avoided division of fractions. This is not hard, in fact, it’s almost as easy as multiplying but I would like to show why the method works.

First consider 6 ÷ 2. This problem is asking the question, “How many 2’s fit into 6?”. The answer is 3 sets of 2’s make up 6. If the problem was 6 ÷ 3, the question would be “How many 3’s fit into 6?”. The answer here is there are 2 sets of 3’s that make up 6. Of course, the answer is not always an integer, Sometimes there are leftover numbers. For example, how many 2’s can make up 7, in other words, 7 ÷ 2. The answer is there are 3 sets of 2’s that can fit into 7 but there will be 1 left over as sets of 2’s do not exactly make up 7. Now remember that fractions are indicating division as well so the above problems are equivalent to

\[
\frac{6}{2}{,}\hspace{0.33em}\hspace{0.33em}\frac{6}{3}{,}\hspace{0.33em}\hspace{0.33em}\frac{7}{2}
\]

 

So what would \[
{1}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{\frac{1}{2}}
\]

mean? It means the same as before: how many one-halves fit into 1? Now the answer is 2 because there are two halves in a whole. What about \[
{2}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{\frac{1}{2}}
\]? Can you see that the answer is 4? Because if there are two things split into halves, then there will be 4 halves.

Let’s now look at \[
{3}\hspace{0.33em}\div\hspace{0.33em}\frac{3}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{\frac{3}{4}}
\]

Looks like there are 4 three-quarters that fit into 3, and that is correct. Let’s look at one more. What about \[
\frac{3}{4}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{\frac{3}{4}}{\frac{1}{4}}
\]. Can you see that there are 3 one-quarters that fit into three-quarters?

All of the above answers can be obtained by multiplying and remembering that integers like 6 can also be shown as a fraction: \[
\frac{6}{1}
\] as 6 divided by 1 is still 6.

So now let’s look at the previous problems. You can convert a fraction division problem by inverting the fraction in the denominator then multiplying:

\[
{1}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{2}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{2}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{2}{1}\hspace{0.33em}{=}\hspace{0.33em}{2}
\]

The inverted fraction is called a reciprocal. A simply way to remember how to do fraction division is the phrase: invert and multiply.

Now let’s do the other ones:

\[
{3}\hspace{0.33em}\div\hspace{0.33em}\frac{3}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{\frac{3}{4}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}}{1}\hspace{0.33em}\times\hspace{0.33em}\frac{4}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}\frac{4}{1}\hspace{0.33em}{=}\hspace{0.33em}{4}
\]

And finally

\[
\frac{3}{4}\hspace{0.33em}\div\hspace{0.33em}\frac{1}{4}\hspace{0.33em}{=}\hspace{0.33em}\frac{\frac{3}{4}}{\frac{1}{4}}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{\rlap{/}{4}}\hspace{0.33em}\times\hspace{0.33em}\frac{\rlap{/}{4}}{1}\hspace{0.33em}{=}\hspace{0.33em}\frac{3}{1}\hspace{0.33em}{=}\hspace{0.33em}{3}
\]

I will do more examples in my next post including division with mixed numbers.