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Can someone please prove the properties of exponents? Doing this will allow me to understand it better. Thanks. 

 Dec 20, 2017

Best Answer 

 #1
avatar+2440 
+3

Sure, I think I can give you a fairly simple proof for you. There are a few property of exponents that you have learned. I have listed all of them for you and what each says. To understand these proofs, you should understand that for whole numbers of a, \(x^a=\underbrace{x*x*x*...*x}\\ \quad\quad\quad\quad\text{a times}\). This is the basis of the notation. The proofs are based on the notation. Sometimes, some rules will assume that a previous one was already proven.

 

Property of Exponent Definition
1. Product of Powers \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=x^{\textcolor{red}{a}+\textcolor{blue}{b}}\)
2. Quotient of Powers \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=x^{\textcolor{red}{a}-\textcolor{blue}{b}}\)
3. Negative Exponents \(x^{\textcolor{blue}{-b}}=\frac{1}{x^\textcolor{blue}{b}}\\ \frac{1}{x^\textcolor{blue}{-b}}=x^{\textcolor{blue}{b}} \)
4. Power of a Power \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}\textcolor{blue}{b}}\)
5. Power of a Product \((xy)^\textcolor{red}{a}=x^\textcolor{red}{a}y^\textcolor{red}{a}\)
6. Power of a Quotient \(\left(\frac{x}{y}\right)^\textcolor{red}{a}=\frac{x^\textcolor{red}{a}}{y^\textcolor{red}{a}}\\ \left(\frac{x}{y}\right)^\textcolor{red}{-a}=\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}\)
   


I will write the proofs in order.

 

1) \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=\underbrace{\underbrace{x*x*x*...*x}*\underbrace{x*x*x..*x}\\ \hspace{7mm}\text{a times}\hspace{18mm}\text{b times}}\\ \hspace{40mm}\text{a+b times}\\ \), so \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}+{\textcolor{blue}{b}}}\)

 

 

2) \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=\frac{\underbrace{x*x*x*...*x}\\ \hspace{2mm}\text{a times}}{\underbrace{x*x*x*...*x}\\ \hspace{2mm}\text{b times}}\). Here, the x in the numerator will cancel out the x in the denominator a-b times, so \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=x^{\textcolor{red}{a}-\textcolor{blue}{b}}\)

 

3a) 

\(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=x^{\textcolor{red}{a}-\textcolor{blue}{b}}\) This was established in the second proof. Set a=0 to get the case of x to the power of -b.
\(\frac{x^0}{x^{\textcolor{blue}{b}}}=x^{0{\textcolor{blue}{-b}}}\) x^0=1, except when x=0.
\(\frac{1}{x^{\textcolor{blue}{b}}}=x^{{\textcolor{blue}{-b}}}\)  
   

 

 

3b) 

\(\frac{1}{x^\textcolor{blue}{-b}}\) As we established in the previous proof, we can replace x^(-b) with 1/[x^(-b)]
\(\frac{1}{\frac{1}{x^\textcolor{blue}{b}}}*\frac{x^\textcolor{blue}{b}}{x^\textcolor{blue}{b}}\) Now, simplify this complex fraction. Notice how the denominators cancel out here.
\(x^\textcolor{blue}{b}\)  
   

 

 

4) \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=\underbrace{x^{\textcolor{red}{a}}*x^{\textcolor{red}{a}}*x^{\textcolor{red}{a}}*...*x^{\textcolor{red}{a}}}= x^{\underbrace{{\textcolor{red}{a}}+{\textcolor{red}{a}}+{\textcolor{red}{a}}+...+{\textcolor{red}{a}}}}\\ \hspace{27mm}\text{b times}\hspace{23mm}\text{b times}\) 

 

If there are b lots of a, this is equivalent to multiplication, so \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}\textcolor{blue}{b}}\)

 

5) \((xy)^\textcolor{red}{a}=\underbrace{xy*xy*xy*...*xy}=\underbrace{(x*x*x*...*x)}*\underbrace{(y*y*y...*y)}=x^\textcolor{red}{a}y^\textcolor{red}{a}\\ \hspace{27mm}\text{a times}\hspace{27mm}\text{a times}\hspace{22mm}\text{a times}\)

 

If we multiply lots of x and y a times, then we can convert this back to an exponent.

 

6a) \(\left(\frac{x}{y}\right)^\textcolor{red}{a}=\underbrace{\frac{x}{y}*\frac{x}{y}*\frac{x}{y}*...*\frac{x}{y}} =\underbrace{(x*x*x*...*x)}*\underbrace{\left(\frac{1}{y}*\frac{1}{y}*\frac{1}{y}*...*\frac{1}{y}\right)} =\frac{x^\textcolor{red}{a}}{1}*\frac{1}{y^\textcolor{red}{a}}=\frac{x^\textcolor{red}{a}}{y^\textcolor{red}{a}}\\ \hspace{27mm}\text{a times}\hspace{22mm}\text{a times}\hspace{25mm}\text{a times}\)

 

6b) 

\(\left(\frac{x}{y}\right)^\textcolor{red}{-a}\) Use the negative exponents rule that we already proved.
\(\frac{1}{\left(\frac{x}{y}\right)^\textcolor{red}{a}}\) Use the rule we used in 6a.
\(\frac{1}{\frac{x^\textcolor{red}{a}}{y^\textcolor{red}{a}}}*\frac{\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}}{\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}}\) Of course, we are only multiplying the fraction by one, which does not actually change the value at all. Now, a lot of canceling occurs in the denominator.
\(\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}\)  
   

 

 Hopefully, these proofs will aide you in your math journey. Always question why something is and don't accept them as fact!

 Dec 21, 2017
 #1
avatar+2440 
+3
Best Answer

Sure, I think I can give you a fairly simple proof for you. There are a few property of exponents that you have learned. I have listed all of them for you and what each says. To understand these proofs, you should understand that for whole numbers of a, \(x^a=\underbrace{x*x*x*...*x}\\ \quad\quad\quad\quad\text{a times}\). This is the basis of the notation. The proofs are based on the notation. Sometimes, some rules will assume that a previous one was already proven.

 

Property of Exponent Definition
1. Product of Powers \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=x^{\textcolor{red}{a}+\textcolor{blue}{b}}\)
2. Quotient of Powers \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=x^{\textcolor{red}{a}-\textcolor{blue}{b}}\)
3. Negative Exponents \(x^{\textcolor{blue}{-b}}=\frac{1}{x^\textcolor{blue}{b}}\\ \frac{1}{x^\textcolor{blue}{-b}}=x^{\textcolor{blue}{b}} \)
4. Power of a Power \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}\textcolor{blue}{b}}\)
5. Power of a Product \((xy)^\textcolor{red}{a}=x^\textcolor{red}{a}y^\textcolor{red}{a}\)
6. Power of a Quotient \(\left(\frac{x}{y}\right)^\textcolor{red}{a}=\frac{x^\textcolor{red}{a}}{y^\textcolor{red}{a}}\\ \left(\frac{x}{y}\right)^\textcolor{red}{-a}=\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}\)
   


I will write the proofs in order.

 

1) \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=\underbrace{\underbrace{x*x*x*...*x}*\underbrace{x*x*x..*x}\\ \hspace{7mm}\text{a times}\hspace{18mm}\text{b times}}\\ \hspace{40mm}\text{a+b times}\\ \), so \(x^{\textcolor{red}{a}}*x^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}+{\textcolor{blue}{b}}}\)

 

 

2) \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=\frac{\underbrace{x*x*x*...*x}\\ \hspace{2mm}\text{a times}}{\underbrace{x*x*x*...*x}\\ \hspace{2mm}\text{b times}}\). Here, the x in the numerator will cancel out the x in the denominator a-b times, so \(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=x^{\textcolor{red}{a}-\textcolor{blue}{b}}\)

 

3a) 

\(\frac{x^{\textcolor{red}{a}}}{x^{\textcolor{blue}{b}}}=x^{\textcolor{red}{a}-\textcolor{blue}{b}}\) This was established in the second proof. Set a=0 to get the case of x to the power of -b.
\(\frac{x^0}{x^{\textcolor{blue}{b}}}=x^{0{\textcolor{blue}{-b}}}\) x^0=1, except when x=0.
\(\frac{1}{x^{\textcolor{blue}{b}}}=x^{{\textcolor{blue}{-b}}}\)  
   

 

 

3b) 

\(\frac{1}{x^\textcolor{blue}{-b}}\) As we established in the previous proof, we can replace x^(-b) with 1/[x^(-b)]
\(\frac{1}{\frac{1}{x^\textcolor{blue}{b}}}*\frac{x^\textcolor{blue}{b}}{x^\textcolor{blue}{b}}\) Now, simplify this complex fraction. Notice how the denominators cancel out here.
\(x^\textcolor{blue}{b}\)  
   

 

 

4) \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=\underbrace{x^{\textcolor{red}{a}}*x^{\textcolor{red}{a}}*x^{\textcolor{red}{a}}*...*x^{\textcolor{red}{a}}}= x^{\underbrace{{\textcolor{red}{a}}+{\textcolor{red}{a}}+{\textcolor{red}{a}}+...+{\textcolor{red}{a}}}}\\ \hspace{27mm}\text{b times}\hspace{23mm}\text{b times}\) 

 

If there are b lots of a, this is equivalent to multiplication, so \(\left(x^{\textcolor{red}{a}}\right)^{\textcolor{blue}{b}}=x^{{\textcolor{red}{a}}\textcolor{blue}{b}}\)

 

5) \((xy)^\textcolor{red}{a}=\underbrace{xy*xy*xy*...*xy}=\underbrace{(x*x*x*...*x)}*\underbrace{(y*y*y...*y)}=x^\textcolor{red}{a}y^\textcolor{red}{a}\\ \hspace{27mm}\text{a times}\hspace{27mm}\text{a times}\hspace{22mm}\text{a times}\)

 

If we multiply lots of x and y a times, then we can convert this back to an exponent.

 

6a) \(\left(\frac{x}{y}\right)^\textcolor{red}{a}=\underbrace{\frac{x}{y}*\frac{x}{y}*\frac{x}{y}*...*\frac{x}{y}} =\underbrace{(x*x*x*...*x)}*\underbrace{\left(\frac{1}{y}*\frac{1}{y}*\frac{1}{y}*...*\frac{1}{y}\right)} =\frac{x^\textcolor{red}{a}}{1}*\frac{1}{y^\textcolor{red}{a}}=\frac{x^\textcolor{red}{a}}{y^\textcolor{red}{a}}\\ \hspace{27mm}\text{a times}\hspace{22mm}\text{a times}\hspace{25mm}\text{a times}\)

 

6b) 

\(\left(\frac{x}{y}\right)^\textcolor{red}{-a}\) Use the negative exponents rule that we already proved.
\(\frac{1}{\left(\frac{x}{y}\right)^\textcolor{red}{a}}\) Use the rule we used in 6a.
\(\frac{1}{\frac{x^\textcolor{red}{a}}{y^\textcolor{red}{a}}}*\frac{\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}}{\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}}\) Of course, we are only multiplying the fraction by one, which does not actually change the value at all. Now, a lot of canceling occurs in the denominator.
\(\frac{y^\textcolor{red}{a}}{x^\textcolor{red}{a}}\)  
   

 

 Hopefully, these proofs will aide you in your math journey. Always question why something is and don't accept them as fact!

TheXSquaredFactor Dec 21, 2017
 #2
avatar
+3

X^2: You would make a great Math teacher if you aren't or weren't one already!!. Congrats on a very thorough job.

 Dec 21, 2017
 #3
avatar+2440 
+1

Well, thank you!

TheXSquaredFactor  Dec 22, 2017

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