My friend Guillermo has recently posted a similar topic on his Mathematics and Multimedia blog.
This concept is actually amazing simple. At first, you might think that there is some long and calculated proof involved. In actuality, this rule for working with negative exponents can easily be found merely by considering the Exponent Laws. (You still remember them, right? Refer back if you need a refresher!) ;)
To keep things simple, I will provide a brief outline that will hopefully describe this concept sufficiently, though if not, I will defer you to Guillermo's page for more details.
Briefly, let's consider the general expression x-n. This can be rewritten to show that the "-n" is being taken away from 0... this is always just assumed whenever we consider a negative number, but write it out this time. So, now we have x0-n. Now, using the Exponent Laws, we know that when we have a subtraction function in the exponent, this is the same as dividing the exponential terms (refer back to my Exponent Laws post, linked above!). And having rearranged things now, you can call upon the property of the zero exponent (see link above!) to show that the numerator is simply just 1, leaving you with the reciprocal base of what we started with. Those few short steps using the Exponent rules demonstrate how x-n is related to 1/xn.
Hopefully that quick walkthrough is a good enough guide to get you through those steps. One thing to take away from this is just how important and useful the Exponent Laws are. Make sure you understand them!
Thanks for visiting Technomaths and your kind words there. I have added your blog in my list of links and I hope you get some visitors to your site. Your blog is a great example of what self-motivated students can find to assist them with their maths work. You have done a grab job explaining some difficult concepts. Well done!
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