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January 10th, 2018, 05:53 AM   #1
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Exponent laws

If anyone is interested, here is a lesson I made on exponent laws with examples. Please leave a comment if you have any question.

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January 10th, 2018, 07:56 AM   #2
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You state the laws of exponents and give some examples but don't explain why those laws are as they are. This is what I have done sometimes:

For h and k positive integers, "$x^k$" simply means "x multiplied by itself k times": $x^k= (x\cdot x\cdot x \cdot\cdot\cdot x)$ (k times).
So $x^kx^h= (x\cdot x\cdot x \cdot\cdot\cdot x)(x\cdot x\cdot x \cdot\cdot\cdot x)$ the first multiplication k times, the second h times. Joining those two sets of "x"s together, there are h+ k "x"s so $x^kx^h=(x\cdot x\cdot x \cdot\cdot\cdot x)$ h+ k times so $x^kx^h= x^{k+ h}$.

Similarly $(x^k)^h$ is just $(x^k)(x^k)\cdot\cdot\cdot (x^k)$ h times so is $(x\cdot x\cdot x \cdot\cdot\cdot x)((x\cdot x\cdot x \cdot\cdot\cdot x)\cdot\cdot\cdot(x\cdot x\cdot x \cdot\cdot\cdot x)$ where we have h parentheses where every parenthesis contains k "x"s. Putting those together, we have hk "x"s multiplied together: $(x^k)^h= x^{kh}$.

That is for h and k positive integers. Now, extend to other sets- because $x^kx^h= x^{k+h}$ and $(x^k)^h= x^{hk}$ are very nice properties, we want to define $x^r$ for r other than a positive integer so that those are still true.

For h= 0 we want to have $x^{k+ 0}= x^kx^0$. Of course, k+ 0= k s that is saying that we want to have $x^{k}= x^kx^0$. As long as $x\ne 0$, $x^k\ne 0$ so we can divide both sides by $x^k$ and have $x^0= 1$. If we want $x^kx^h= x^{k+h}$ even when h or k are 0, we must define $x^0= 1$. And you can see why we must restrict x to not be 0 in that case.

Now, suppose h is negative. Say, h= -k. Then $x^{k+ -k}= x^0= 1$. In order to have $x^{h+ k}= x^hx^k$ even in the case that h= -k we must have $x^{k+ -k}= 1= x^kx^{-k}$. Again, as long as x is not 0, we can divide sides by $x^k$ and have $x^{-k}= \frac{1}{x^k}$. In order that $x^kx^h= x^{k+h}$ be true even for negative exponents we must define $x^{-k}= \frac{1}{k}$.

For rational exponents, $x^{\frac{h}{k}}$, we start by looking at $x^{\frac{1}{k}}$ using the other exponent law $x^{kh}= (x^k)^h$. In order that this be true even in the case that $h= \frac{1}{k}$ we must have $x= x^{\frac{k}{k}}= \left(x^{1/k}\right)^k$. Taking the kth root of both sides, we have $x^{1/k}= \sqrt[k]{x}%$. The general rational case follows from the law $x^{hk}= (x^h)^k$: $x^{h/k}= x^{h(1/k)}= (x^h)^{1/k}= \sqrt[k]{x^h}$.

The set of all real numbers is not defined "algebraically", it is defined "analytically"- If r is any real number then there exist a sequence of rational numbers, $\{a_n\}$, that converges to r: $\lim a_n = r$. We use that to define $x^r= \lim x^{a_n}$. For example, $\pi$ is irrational but is the limit of the sequence {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, 3.1415926, ...} (that's what we mean when we say that $\pi$ can be approximated, to any desired accuracy, by "3.1415926...") so that we define $x^{\pi}$ to be the limit of the sequence $\{x^3, x^{3.1}, x^{3.14}, x^{3.141}, x^{3.1415}, x^{3.14159}, x^{3.141592}, ...\}$. That is, we can approximate $x^{\pi}$ to any desired accuracy by taking x to the appropriate approximation of $\pi$.

Last edited by Country Boy; January 10th, 2018 at 07:59 AM.
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