# Algebra Rules for Arithmetic

This is the *distributive property of multiplication*. If you're multiplying something with a sum of two or more other terms, you can *distribute* your multiplication to each of the terms. This can be helpful for simplifying problems.

Any multiplication or division of the numerator of a fraction applies to the fraction as a whole, and vice versa: if you need to multiply a fraction, multiply the numerator and your aim is achieved

Because of the reciprocal relationship of a fraction's numerator and denominator, if you *divide* the numerator by a particular number, it has the same effect on the fraction's overall value as if you *multiply* the denominator by that same number

In the same fashion as the above rule, *dividing* the denominator of a fraction has the same effect as *multiplying* the numerator

This follows from the fact that you can find a common denominator between two fractions by multiplying the numerator and denominator of each fraction by the other's denominator, like so:

This is another version of rule 5, but for *subtraction* of two fractions, rather than addition.

Reversing the order of a subtraction operation creates the inverse result: **5-3 = 2; 3-5 = -2**. A consequence of this is that for a fraction in which both the numerator and denominator contain a subtraction (ie, they each contain a negative term), if both those subtractions are reversed, the value of the fraction remains the same. This is because *if the sign of both the numerator and denominator of a fraction is changed, the fraction's value remains the same*.

This is simply a result of the fact that two fractions with common denominators can be added by adding the numerators and leaving the denominator unchanged. It is intuitively simple: 1 third plus 2 thirds is 3 thirds. The reverse is also true: 3 thirds minus 2 thirds equals 1 third.

Division can be thought of as the inverse of multiplication: if ``{a \over b} = c`` then ``b*c = a``. This means that the fraction ``{ac \over c}`` is equal to ``a``, since we are multiplying ``a`` by ``c`` and then immediately dividing it by ``c`` again, which puts us right back to where we started. This rule is an extension of that fact. Since we know that ``{ac+bc \over c} = {ac \over c} + {bc \over c}`` (see rule 8), and based on the above we can see that ``{ac \over c} = a`` and ``{bc \over c} = b``, from that we have our result: ``a+b``.

This is a combination of rule 3 and rule 4.

# Algebra Rules for Exponents

As with many rules related to exponents, writing out the exponents as multiplications makes it obvious why the rule is true

Like the previous rule, this one can be demonstrated simply by expanding the exponents out into a series of multiplications

Thanks to the commutative property of multiplication, any series of multiplications can be rearranged without changing its value. This means that we can take a multiplication raised to a power and rearrange the resulting series of multiplications to make two exponents

It might seem odd to have a negative exponent (since you can't multiply something by itself a negative number of times). However, if we take a closer look at the rule ``a^na^m = a^{n+m}`` we can see that it implies that ``a^{-n}`` must equal ``{1 \over a^n}``, the *multiplicative inverse* or *reciprocal* of ``a^n``.

This becomes clear looking at the ``a^{n+m}`` side of the equation from rule 11. What happens if ``m`` is negative? Obviously, this will reduce the combined value of the exponent (for example, ``2^{4-2} = 2^2``). What does this mean for the *left* hand side of the ``a^na^m = a^{n+m}`` equation? It means that the value of, for example, ``2^4`` must be reduced to ``2^2`` when it is multiplied by ``2^{-2}``. If, as this rule states, ``a^{-n} = {1 \over a^n}``, this works out perfectly: ``2^4 * 2^{-2} = 2^4 * {1 \over 2^2} = 16 * {1 \over 4} = 4 = 2^2 = 2^{4-2}``

The reciprocal of a fraction is the fraction turned on its head: the reciprocal of ``{2 \over 3}`` is ``{3 \over 2}``. We know from the previous rule that ``a^{-n}`` is the reciprocal of ``a^n``, so we can simply convert the fraction to its reciprocal by exchanging the numerator and denominator, and then the exponent becomes positive. Positivity is such a nice thing!

This looks weird at first, but the reasons behind it are pretty simple. If we were paying attention when someone told us how to multiply fractions (this is doubtful, but we'll continue anyhow) we will remember that to multiply two fractions you simply multiply the numerators with each other and multiply the denominators with each other to get the resulting fraction. This rule follows from that fact.

This one is very simple. Since division is the inverse of multiplication, multiplying a number by itself a few times and then dividing it by itself multiplied a few time is the same as just multiplying it by itself *a few less times*.

This rule may seem arbitrary, but it is necessary in order to maintain consistency with other properties of exponents. Consider the rule ``a^na^m = a^{n+m}``. What happens if ``m = 0``? The right hand side of the equation will be ``a^{n+0}``, or ``a^n``. This means that in the left hand side, ``a^n`` has to be multiplied by the value of ``a^0``, but remain unchanged. The only way for this to be the case is if ``a^0 = 1``. (For some discussion of the peculiar case of ``0^0`` and why it should (probably) equal ``1``, see this article.)

# Algebra Rules for Radicals

As in some of the exponent properties, this rule is not an intuitive extension of the typical meaning of an exponent. Nevertheless, it fits with the all-important exponent rule ``a^na^m = a^{n+m}``.

If ``x = \sqrt{a}`` and ``y = \sqrt{b}`` then ``\sqrt{ab} = \sqrt{x^2*y^2}`` If we write out the multiplication, this turns into ``\sqrt{x*x*y*y}``. Thanks to the commutative property of multiplication, we can rearrange the Xs and Ys and get ``\sqrt{x*y*x*y} = \sqrt{(x*y)(x*y)} = \sqrt{(x*y)^2} = x*y = \sqrt{a}*\sqrt{b}``

Once again, by working backwards from the value of these two expressions we can see why they are equal. If ``\sqrt[m]{\sqrt[n]{a}} = x``, then we can construct ``a`` out of combinations of ``x`` and see how the whole equation works. To make things simple, we'll start with given values of ``m`` and ``n``. If ``\sqrt[2]{\sqrt[3]{a}} = x``, then ``x = \sqrt[2]{x*x}``, which also means that ``\sqrt[2]{\sqrt[3]{(x*x)*(x*x)*(x*x)}} = x``. So ``a = (x*x)*(x*x)*(x*x) = x^6 = x^{mn}``. And happily, ``\sqrt[mn]{x^{mn}} = x`` by definition, so we have ``\sqrt[m]{\sqrt[n]{a}} = x = \sqrt[mn]{x^{mn}}``. Our example is a specific case where ``m = 2`` and ``n = 3``, but since ``a`` will always be equal to ``x^{mn}``, the equation holds regardless of the values of ``m`` and ``n``

To see how this works, we can use a similar trick to the rule above. If ``x = \sqrt{a} `` and ``y = \sqrt{b}``, then: ``\sqrt{{a \over b}} = \sqrt{{x*x \over y*y}} = \sqrt{{x \over y}*{x \over y}} = \sqrt{{x \over y}^2} = {x \over y} = {\sqrt{a} \over \sqrt{b}}``

If ``a`` is a positive number, then ``\sqrt[n]{a^n}`` will always equal ``a``. However, if ``a`` is negative, then the result will be positive if ``n`` is even, but it will be negative if ``n`` is odd. This comes from the fact that multiplying a negative number an even number of times always produces a positive result; an odd number of multiplications will produce a negative result. (This has the interesting side effect that there are no real numbers that are even-numbered roots of a negative number.)