On 24 Jul 2009, at 14:20, m.a. wrote: > Bruno, > You overlooked the question at the bottom of the page > that I tried unsuccessfully to work out. Brent supplied the answer > but what I was looking for were the steps that lead up to the > answer. marty > > > ----- Original Message ----- > From: Bruno Marchal > To: everything-list@googlegroups.com > Sent: Friday, July 24, 2009 4:48 AM > Subject: Re: Seven Step Series > >> >> a^n * a^(-m) = a^(n-m) >> >> Again, verify this on simple example of your own. >> >> OK: If a=10 and n=3 and m=4 Following the formula above >> "a^n * a^(-m)" ,I get as the first half of the equation: >> >> 10^3 * 10/4 =1000 x 2.5= 2500 but for the second half "a^(n- >> m) I get:
"first half": It is 10^3 * 10^(-4) . OK? Now, it looks like you are saying that 10^(-4) is 10/4. But we have defined a^(-n) by 1/a^n. So 10^(-4) is 1/(10^4) = 1/10000 so it is 1000 * 1/10000 = 1000/10000 = 1/10 = 0.1 >> >> >> 10^(n-m)= 10^ -1= 10/1 >> >> which of course makes no sense at all. Where did I go wrong? "second half": 10^(n-m) = 10^(3-4) = 10^(-1) = 1/(10^1) = 1/10 = 0.1 (you found 10/1, which is 10). My diagnostic: You have not integrate that a^(-n) is: ONE divided by a^n. It is 1 / a^n. So 10^(-4) = 1/10^4 = 1/10000 = 0.0001 and 10^(-1) = 1/10^1 = 1/10 = 0.1 Be careful with the little numbers 10^1 = 10, 10^0 = 1. (see below if you have a problem) 1 divided by any number bigger than 1 is always a number little than 1. With decimal they begin by 0.<something>. For example, when we write a rational number like 234.567 it is an abbreviation of 2*100 + 3*10 + 4*1 + 5*(1/10) + 6*(1/100) + 7*(1/1000), which is the same as: 2*(10^2) + 3*(10^1) + 4*(10^0) + 5*10^(-1) + 6*10^(-2) + 7*10^(-3) We say that 234.567 is written in base ten. All the digits are coefficient of power of ten. (where a power of 10 is a number = 10^n, n any integer). Are you OK with 10^0 = 1? If not read below 1 = 100/100 OK? 100/100 = 10^2/10^2 OK? 10^2/10^2 = 10^(2-2) OK? (use of the formula above) 10^(2-2) = 10^0 OK? So: 1 = 10^0 OK? So 10^0 = 1 OK? Question? I think you could have found the mistakes by carefully reread the definitions, in this case, of a^(-n), which is 1 / (a^n). Don't you think so? Bruno http://iridia.ulb.ac.be/~marchal/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to everything-list@googlegroups.com To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---
