Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Why a Zero Exponent Equals One, and Changing Number Bases


Date: 9/26/95 at 19:24:48
From: Anonymous
Subject: Business 

Dear Dr. Math:

Why is any number to the zero power equal to one?

Also, could I have some information on hexadecimal and 
binary for my classes.

Thank you!

Carol Lyons


Date: 9/30/95 at 15:52:21
From: Doctor Ken
Subject: Re: Business 

Hello!

About your first question:  I think the right way to think 
about exponents is to say that a^k is equal to 1 times k x's.  
So 2^3 = 1*2*2*2 = 8, 2^(-4) = 1*1/2*1/2*1/2*1/2 = 1/16 since 
multiplying something once should be the opposite of dividing 
something once.  With this concept in mind, any number to the 
zero power is "1 times no repetitions of that number."  So 
any number to the zero power is 1.

Also, look at the following chart:

   2^5      2^4     2^3      2^2      2^1      2^0     2^-1     2^-2
    32       16      8        4        2        1       1/2      1/4

Every time you go one higher in the exponent, you multiply by 
2 (and every time you go one lower, you divide by 2).

Here's how hexadecimals and binaries work.  What we usually 
deal with is base ten.  It's just a method of notation, a way 
that's convenient for us to write down numbers.  It means 
that if we have the number 9745, that's 9*10^3 + 7*10^2 + 
4*10^1 + 5*10^0.

If there are numbers after the decimal point, you just 
continue the pattern:

234.95 = 2*10^2 + 3*10^1 + 4*10^0 + 9*10^-1 + 5*10^-2

So it's actually very related to your first question.

The only difference between base 10 and base anything else is 
that we replace 10 (as in 9*10^3) with the new number, and 
instead of using 10 digits, we now use however many digits 
our base is.  So in base 2 (binary) we use 2 digits, 0 and 1, 
and in base 16 (hexadecimal) we use 16 digits, 
0,1,2,3,4,5,6,7,8,9,a,b,c,d,e, and f.

To convert the number 984 to hexadecimal, we'd try to write 
it as w*16^3 + x*16^2 + y*16^1 + z*16^0 where w,x,y, and z 
are between 0 and 15(f).

Since 16^3 = 4096, we have w=0, since 1 would be too big.

16^2 = 256.  How many times does 256 go into 984?  Three.  
That's x.  So now we have 984 = 3*16^2 + y*16^1 + z*16^0.  
So 216 = y*16^1 + z*16^0.  

Since 16 goes into 216 13 times, y is 13, i.e. "d".  So we have
     984 = 3*16^2 + 13*16^1 + z*16^0.  So
       8 = z*16^0, and z=8.

So 984 in hexadecimal is 3d8.

How would you write 984 in binary?
      
-Doctor Ken,  The Geometry Forum

    
Associated Topics:
High School Number Theory

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/