+0

# Simple Proofs (Hopefully)

0
70
3
+1601

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

TheXSquaredFactor  Dec 20, 2017

#1
+1601
+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!

TheXSquaredFactor  Dec 21, 2017
Sort:

#1
+1601
+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!

TheXSquaredFactor  Dec 21, 2017
#2
+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.

Guest Dec 21, 2017
#3
+1601
+1

Well, thank you!

TheXSquaredFactor  Dec 22, 2017

### 24 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details