I have taught algebra many times, and one thing that is amazing is that exponents are presented several times. There is a need for this. Constraints happen as we expand out studies past positive integers, yet students often study as though the whole concept is new.

The rules for exponents are the same when they are presented numerus times in a textbook, except one cannot divide by 0, one is not to raise 0 to the 0-th power, and one cannot take as even root of a negative number in the real number system. An even root occurs when we have a fractional power with an even number for the denominator, such as 5^{1/2}. When one of the difficulties mentioned occurs, an author restates the same rules. Learn them once and allow the authors to restate them.

In other words, when the rules for exponents are first encountered it may be for positive integers. When this is extended to include the number zero, such as when all integers are considered, we must be careful to not divide by zero or have zero raised to the zeroth power. Just follow the constraints of algebra and the rules learned once can still be used, with no fear something different is being added.

Let’s go through exponents once, learn the rules, and apply constraints where appropriate. We will understand by example, which is not technically a proof. Our examples will involve integers, but the same rules apply for fractions, and even complex numbers.

*a*^{n}a^{m} = *a*^{n}^{+m}

8^{2}8^{4} = 8^{2 + 4} = 8^{6}

*a*^{n}/*a*^{m} = *a*^{n }^{– m}

7^{5}/7^{2} = 7^{5 – 2} = 7^{3}

*a*^{0} = 1, if *a* ≠ 0

*a*^{n}/*a*^{n} = 1, and the above rule makes this the same as *a*^{0}.

*a*^{-n} = 1/*a*^{n}

This is a definition. It is equally correct to say *a*^{n} = 1/*a*^{-n}.

2^{-3} = 1/2^{3}

(*a*^{n})^{m} = *a*^{nm}

(5^{2})^{3} = 5^{2X3}

(5^{2})^{3} = (5^{2})(5^{2})(5^{2}) = (5X5)(5X5)(5X5) = 5^{6}

5^{2X3} = 5^{6}

(*ab*)^{n}=*a*^{n}b^{n}

(4X3)^{2} = 4X3X4X3 = 4^{2}3^{2}

(*a/b*)^{n} = *a*^{n}b^{n}

(2/3)^{2} = (2/3)(2/3) = 2^{2}/3^{2}

Finally, we get the last two, which can be gotten from the above.

(*ab*)^{n}(*ab*)^{m} = *a*^{n}^{ + m}*b*^{n }^{+ m} and (*a*/*b*)^{n}(*a*/*b*)^{m} = *a*^{n}^{ + m}/*b*^{n }^{+ m}

## Comments

Loginblackspanielgallery on 05/03/2021Simple reviews that do not bog down in complexities unique to one problem can refresh what has been learned.

WriterArtist on 05/03/2021It is refreshing to go through Algebra I thought was forgotten. Fortunately I found that I remembered everything when I read your article.

blackspanielgallery on 02/12/2021I have taught algebra so many times I just make them up as I go. Experience allows me to know the variety of examples needed, and I try for a good coverage.

DerdriuMarriner on 02/12/2021blackspanielgallery, Thank you for the practicalities and the products.

How do you choose the examples that you did here and in your other algebra series articles?