June 6th, 2014, 12:00 AM  #1 
Senior Member Joined: Oct 2012 From: Live and work in Saudi but from Italy (British national) Posts: 103 Thanks: 2 Math Focus: Hoping to do something in the future.  x^2 = 1/x^2
Hi Can someone point me to a clear explanation of why x^2 = 1/x^2 I know that the rule is that negative exponents move to the other side of the fraction and become positive exponents. But why? Thanks Mike 
June 6th, 2014, 01:52 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,150 Thanks: 730 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Multiply both sides of the equation by$\displaystyle x^2$... 
June 6th, 2014, 05:40 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra 
It's more or less a definition, although a natural one. We have $x^{n} = x \cdot x^{n1}$. Satisfy yourself that this is true by putting some positive integers in place of $n$. Now, dividing both sides by $x$ we get $$\frac{x^n}{x} = x^{n1}$$ From this we can work down: \begin{align*} \frac{x^2}{x} &= x^1 = x \\ \frac{x^1}{x} &= x^0 = 1 \\ \frac{x^0}{x} &= x^{1} = \frac{1}{x} \\ \frac{x^{1}}{x} &= x^{2} = \frac{1}{x^2} \\ &\cdots \end{align*} Last edited by skipjack; June 6th, 2014 at 06:32 AM. 
June 6th, 2014, 05:41 AM  #4 
Member Joined: Feb 2012 From: Hastings, England Posts: 83 Thanks: 14 Math Focus: Problem Solving 
I like to look at what happens to reducing powers of x. if we write it out as: $\displaystyle x^5 = x.x.x.x.x$ $\displaystyle x^4 = x.x.x.x$ $\displaystyle x^3 = x.x.x$ $\displaystyle x^2 = x.x$ $\displaystyle x^1 = x$ We can see that to reduce the power each time we divide by x. Hence why $\displaystyle x^0 = 1$ as this is $\displaystyle x^1 / x = x/x = 1$ then we keep going $\displaystyle x^{1} = 1/x$ $\displaystyle x^{2} = 1/x.x = 1/x^2$ Hope that is clear enough to follow PS: Don't know why it wont show my negative powers properly still new to this stuff Last edited by nmenumber1; June 6th, 2014 at 05:57 AM. 
June 6th, 2014, 05:41 AM  #5 
Member Joined: Feb 2012 From: Hastings, England Posts: 83 Thanks: 14 Math Focus: Problem Solving 
this keeps happening to me, finish posting and by them someones already there with a better and neater explanation lol good job mate

June 6th, 2014, 05:51 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra  
June 6th, 2014, 05:56 AM  #7 
Member Joined: Feb 2012 From: Hastings, England Posts: 83 Thanks: 14 Math Focus: Problem Solving 
ah nice one thanks man 