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Why is it that if you put 2^0 you would get 1?

 Jun 3, 2017
 #1
avatar+9460 
+1

Any number, except for zero, raised to the power of zero equals 1 .

 

 

Here is one way of thinking about why this is. I've never heard it explained like this, but I really like his explanation! I will probably think of it this way from now on!

https://www.youtube.com/watch?v=dAvosUEUH6I

 

 

And....I was trying to find a good video explaining it another way, but I couldn't really find a good one. Here's a pretty good answer to your question that you can read through if you want.

http://mathforum.org/dr.math/faq/faq.number.to.0power.html

 Jun 3, 2017
edited by hectictar  Jun 3, 2017
 #2
avatar+2439 
+2

I think I have an explanation that is simple to understand! In advance, I have not referenced any of Hecticlar's sources before writing this explanation, so mine could be similar or different. To start, I am going to make a table of powers that we can calculate:

 

\(2^n\) Written-out Result
\(2^5\) 2*2*2*2*2 32
\(2^4\) 2*2*2*2 16
\(2^3\) 2*2*2 8
\(2^2\) 2*2 4
\(2^1\) 2 2
\(2^0\) ? ?
\(2^{-1}\) 1/(2) 1/2
\(2^{-2}\) 1/(2*2) 1/4
\(2^{-3}\) 1/(2*2*2) 1/8
\(2^{-4}\) 1/(2*2*2*2) 1/16
\(2^{-5}\) 1/(2*2*2*2*2) 1/32

 

Do you notice a pattern? I do. As you go down the list, you can divide by 2 to get the next number in the sequence! FIrst, I'll generalize this statement:
 

\(\frac{2^n}{2}=2^{n-1}\)

 

What I have done here is manipulate the powers so that I can circumvent raising to the power of 0. If I make n=1, I will raise to the power of zero and get a result of what that answer should be. Let's try it!

 

\(\frac{2^1}{2}=2^{1-1}\) Let's simplify the right hand side first by doing 1-1
\(\frac{2^1}{2}=2^0\) Woah! Evaluate the left hand side to figure out what 2^0 truly equals.
\(\frac{2}{2}=2^0\)  
\(1=2^0\)  
   

 

We can generalize this further to say that any number raised to the power of zero is 1 using some algebra:

 

\(1=\frac{x^n}{x^n}\hspace{1mm},x\neq0\) This statement is true because any number divided by itself is one! I'll use an exponent rule that says that \(\frac{x^n}{x^n}=x^{n-n}\)
\(1=x^{n-n}\) n-n=0, so let's simplify that
\(1=x^0\hspace{1mm},x\neq0\) This is saying that any number to the power of zero is one. 
   
 Jun 3, 2017
 #3
avatar+4609 
0

Guest, there's 1 basic rule. Anything to the power of 0, is one. So, \(2^0\)=1

 Jun 3, 2017

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