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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 January 10th, 2018, 04:53 AM #1 Member   Joined: Oct 2017 From: Japan Posts: 62 Thanks: 3 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. January 10th, 2018, 06:56 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 06:59 AM. Tags exponent, laws Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post PGibson Algebra 10 August 20th, 2017 09:45 PM BreakingTheCode Algebra 1 March 18th, 2014 08:57 PM FloorPlay Applied Math 2 December 4th, 2012 12:11 PM happysmiles374 Elementary Math 2 February 28th, 2012 08:48 AM dathomiedre Algebra 1 January 25th, 2010 02:17 PM

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