We can simplify same numbers multiplied together multiple times into indices or (to the power of numbers) and we are then able to find what number multiplied by itself (a certain number of times) makes the number by using roots.
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A square number is an integer that 2 of the same number multiply together to make. For example:
$$ 2 \times 2 = 2^{2} = 4 $$
4 is a square number because 2 numbers square to make it. The reason these numbers are called “square numbers” is because you calculate the area of a square by multiplying 2 sides. cube numbers are similar to square numbers but involve 3 of the same number (the volume of a cube).
roots are the inverse of indices. In other words to find the original number after “squaring” it (multiplying it by itself) you would have to find the square root of the product of the number squared. For example:
$$ 3 \times 3 = 3^{2} = 9 $$ $$ \sqrt{9} = 3 $$
roots like powers, can apply to more than just 2 times the number, for example the cube root is 3 lots of the same number that multiply together to make the number.
index notation is when instead of writing:
$$ 2 \times 2 \times 2 \times 2 $$
we write $2^{4} $$ where 2 is the base and 4 is the exponent (or the number of times…). There are a few special rules you should be made aware of when dealing with numbers with exponents:
Firstly, when multiplying exponents where the base is the same number, we just add the exponents together and keep the base the same. For example:
$$ 4^3 \times 4^2 = 4^{3 + 2} = 4^5 $$
We can check that this is indeed the case by writing out the lots of the number 4 being multiplied and counting how many there are:
$$ 4 \times 4 \times 4 \times 4 \times 4 = 4^{5} $$
When dividing exponents where the base is the same number we can subtract the exponents and keep the same base:
$$ \dfrac{4^3}{4^2} = 4^{3 - 2} = 4^1 = 4 $$
Now we could write out all the multiplications and cancel out to check that this works but you will just have to trust that it does indeed work.
The above example brings up another rule where when the exponent is 1 then it is just the number (as it’s just 1 lot of the number).
What does a negative exponent mean? Well for some really wierd reason a negative exponent means 1 divided by the number to the positive exponent:
$$ 2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4} $$
By knowing this rule and the above rules for multiplication and division of exponents we can tackle multiplying or dividing numbers with the same base and negative exponents:
$$ 2^{-3} \div 2^{-2} = \dfrac{1}{2^3} \div \dfrac{1}{2^2} = \dfrac{1}{8} \times \dfrac{4}{1} = \dfrac{4 \color{red}{\div 4}}{8 \color{red}{\div 4}} = \dfrac{1}{2}$$ $$ \text{ or: } = 2^{-3 - (-2)} = 2^{-1} = \dfrac{1}{2} $$
It’s probably faster and easier to add or subtract the exponents together then convert to a fraction for the final answer if you’re asked to do so.
Another confusing rule of exponents (also known as indices (plural of index)) is that a number to the power of 0 is always equal to 1:
$$ 10000^0 = 1 $$
Another rule for exponents is that we multiply exponents together when there is a base and an exponent that is surrounded by brackets like this:
$$ (3^2)^4 = 3^{2\color{red}{\times 4}} = 3^8 $$
If there is more than one number (or base and exponent) in a bracket we must do the same to all the numbers. So for example:
$$ (3^2 + 2^3 + 2)^2 = 3^{2 \color{red}{\times 2}} + 2^{3 \color{red}{\times 2}} + 2^{1 \color{red}{\times 2}} = 3^4 + 2^6 + 2^2 $$
Now that we have covered most of the rules of exponents we should continue on from understanding about prime factors. All numbers can be expressed as a series of prime factors multiplied together (or the product of prime factors). It is helpful to express numbers this way as it helps us find the lowest common multiple which is helpful for finding common denominators for adding or subtracting fractions and the highest common factor which is helpful when factorising numbers.
Firstly to express numbers as products of prime factors we just break the number into multiple numbers multiplied up to make the original number. When we can no longer do this we have found prime numbers that won’t brake up anymore. For example let’s express 720 as products of prime factors:
$$ 720 = \color{red}{10 \times 72}= \color{red}{5 \times 2} \times \color{red}{36 \times 2} = 5 \times 2 \times \color{red}{18 \times 2} \times 2 $$ $$ 5 \times 2 \times \color{red}{6 \times 3} \times 2 \times 2 = 5 \times 2 \times \color{red}{3 \times 2} \times 3 \times 2 \times 2 $$ $$ = 2^{3} \times 3^{2} \times 5 $$
We can now find the highest common factor for 720, 45 and 230 if we express all the numbers as a product of their prime factors and then multiply the common prime factors together we can find the highest common factor:
$$ 720 = 2^{3} \times 3^{2} \times 5 $$ $$ 45 = 3^2 \times 5 $$ $$ 230 = 2 \times 5 \times 23 $$
$$ \color{red}{\text{HCF: } 5} $$
The highest common factor is 5 since no other number is in all of the lists. We can also find the lowest common multiple by multiplying all of the prime factors together; however if 1 is in more than one list we list it the most amount of times it appears once in a list:
$$ \color{red}{\text{LCM: } 2^3 \times 3^2 \times 5 \times 23 = 8280} $$
Surds are roots of numbers which are irrational (the number continues forever without stopping or repeating). In other words, the only number that will multiply by itself a certain number of times to make the root of the number must continue forever and be irrational.
There are some rules we need to understand for surds or roots more in general when dealing with them. Firstly, this basic property really reflects what a root of something actually is. If you take a square root of a number and square it you remove the square root of the number and are left with the number, or if you take the nth root number and multiply it nth many times you get the number:
$$ \sqrt{2}^2 = \sqrt{2} \times \sqrt{2} = 2 $$ $$ \sqrt[4]{3} = \sqrt[4]{3} \times \sqrt[4]{3} \times \sqrt[4]{3} \times \sqrt[4]{3} = 3 $$
Next let’s connect what a root is to our understanding of exponents. The square root of a number is the same as the number to the power of a half:
$$ \sqrt{9} = 9^{\tfrac{1}{2}} = 3 $$
Or more generally, the nth root of a number is the same as the number to the power of 1 divided by the nth number (n can be any number). Here is an example:
$$ \sqrt[3]{4} = 4^{\tfrac{1}{3}} $$
Using the above rules we can now make a rule for fractions where the numerator is not 1:
$$ 3^{\tfrac{3}{4}} = 3^{(\tfrac{1}{4} \times \tfrac{3}{1})} = (\sqrt[4]{3})^3 $$
So we can split the number up to find the root of the number and then multiply that by the numerator of the first fraction.
Now that we have related roots and exponents together we can dive into learning the rules of roots (or surds).
$$ 2\sqrt{2} = 2 \times \sqrt{2} $$
The fact that you need not put a multiplication symbol in front of a number to suggest “2 lots of the square root of 2” should remind you of basic algebra (if you have indeed studied it yet!). You can add or subtract roots easily when you have the same root; again like in algebra:
$$ 3 \sqrt{2} - 2 \sqrt{2} = (3 - 2) \sqrt{2} = 1 \sqrt{2} = \sqrt{2} $$
We can also break up roots into factors, this allows us to simplify root equations easily. For example let’s start off with an easy example:
$$ \sqrt{8} = \sqrt{ \color{red}{2 \times 4}} = \color{red}{\sqrt{2} \times \sqrt{2} \times \sqrt{2}} = 2 \sqrt{2} $$
Now we can apply this to an equation:
$$ \sqrt{8} + 3 \sqrt{32} = \color{red}{2 \sqrt{2}} + 3 \sqrt{\color{red}{16 \times 2}} = 2 \sqrt{2} + (3 \sqrt{2} + \color{red}{\sqrt{4} \times \sqrt{4}}) = 2 \sqrt{2} + (3 \sqrt{2} \times \color{red}{4}) $$ $$ 2 \sqrt{2} + \color{red}{12} \sqrt{2} = 14 \sqrt{2} $$
We can now use these rules and apply it to surds equations to solve them really easily. Finally, however, we need to be able to rationalise the denominator whereby we take surds that are in a fraction and try and make the denominator a normal, “rational” number (a number that isn’t irrational). Firstly, we can do this by applying the first rule we learned about surds:
$$ \dfrac{2 \color{red}{\times \sqrt{8}}}{\sqrt{8} \color{red}{\times \sqrt{8}}} = \dfrac{2 \sqrt{8}}{8} $$
We could then simplify this further, but time is of the essence and we have demonstrated rationalising the denominator for a single root. In order to learn how to rationalise the denominator for harder equations with roots, we need to learn about brackets. This is very similar to the rule of exponents and brackets.
With double brackets, we must multiply everything in one bracket by everything in the other bracket. For example:
$$ (1 + 3)(1 + 4) = 1 \color{red}{\times 1} + 3 \color{red}{\times 1} + 1 \color{red}{\times 4} + 3 \color{red}{\times 4} $$
A great way, to check you have mulitiplied everything together is:
An interested property that comes from dealing with “double brackets” (2 brackets squashed together) is known as the difference of 2 squares:
$$ 4^2 - 3^2 = (4 + 3)(4 - 3) $$
This let’s us factorise and solve certain equations really easily. We can use the difference of 2 squares to rationalise the denominator along with understanding that squaring the square root of a number gives us that number:
$$ \dfrac{1}{2 - \sqrt{3}} \color{red}{\times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}}} = \dfrac{2 + \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \dfrac{2 + \sqrt{3}}{2^2 - 3} = 2 + \sqrt{3} $$
So, in simple terms, we multiply by the equation with the opposite symbol in order to get the difference of 2 squares for the denominator.