Hi Kim, Marty, and others On 23 Jul 2009, at 15:48, m.a. wrote:

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> > Yes, this is all clear to me. But when I try to > put it into practice, confusion reigns. (See bottom of page.) By the > way, because I'm printing out your lessons and often need to refer > back to specific ones, it becomes confusing when you append lessons > to posts with different subjects. Thanks, marty a. No problem. Actually I did this because I thought I would be disconnected from the net for some period, so I give some more material. I may come back on this. Also, I work in a spiraling way, and I give a little more than strictly needed for the seventh step, so don't worry if some points remain unclear. It is part of the math experience that we have to come back regularly on the notions, to develop the familiarization. Tell me if most things are OK? I will make a little summary, and move on on the last lesson on set, up to Cantor theorem. After this we move on the "mathematical" notion of computation. Kim, are you able to follow? I feel Marty is quite quick, in quasi hurry. Don't hesitate to ask question and slow us done. Read carefully the next summary, it will be an opportunity to make revision, Best to all of you, Bruno > > > ----- Original Message ----- > From: Bruno Marchal > To: everything-list@googlegroups.com > Sent: Thursday, July 23, 2009 6:37 AM > Subject: Re: Seven Step Series > > > On 23 Jul 2009, at 05:44, m.a. wrote: > > >> >> >> if a is a number, usually, a^n is the result of effectuating (a >> >> times a >> >> times a time a ... times a), with n occurences of a. For example: >> >> 2^3 = >> >> 2x2x2 = 8. >> >> >> >> so a^n times a^m is equal to a^(n+m) >> >> >> >> This extends to the rational by defining a^(-n) by 1/a^n. In that >> >> case >> >> a^(m-n) = a^m/a^n. In particular a^m/a^m = 1 (x/x = 1 always), and >> >> a^m/a^m = a^(m-m) = a^0. So a^0 = 1. So in particular 2^0 = 1. >> >> Do you really expect us to understand this? > > > It would help me if you could be more specific on what is it that > you don't understand. Let me copy the text above, and show how you > could perhaps be more specific. > > I first said: > >> >> >> if a is a number, usually, a^n is the result of effectuating (a >> >> times a >> >> times a time a ... times a), with n occurences of a. For example: >> >> 2^3 = >> >> 2x2x2 = 8. > > > Do you understand this? Can you compute, with the help of a pocket > computer the following: > > 2^4 = ? (answer: 16) > 4^3 = (answer: 64) > 10^5 = (answer: 100000) > > Do you understand a sentence like, "let us assume that a and b are > positive integers (= natural numbers) then a^b = a*a* ... *a with b > occurrences of a."? > > Then I said that > >> >> so a^n times a^m is equal to a^(n+m) > > > I assume here that a, n, and m are positive integers. Have you a > problem here? > > You can always verify such a statement on examples. You should think > like this. > > Bruno pretends that a^n times a^m is equal to a^(n+m). This seems > weird. I guess Bruno means this to be true for all (natural) numbers > a, n and m. > Let us see what this could mean on some little numbers (not too > little because 0, 1 and even 2 are sometimes more complex than > little NUMBERS like 3, 4, 5 ...). > > So let us see with a =3, n = 4, and m = 5. > > The general statement a^n times a^m = a^(n+m) becomes > > 3^4 times 3^5 = 3^(4 + 5) > > Is that true? > > 1) brute force verification: (I wrote "*" as a shorthand for "times") > > 3^4 = 3*3*3*3 = 81 > 3^5 = 3*3*3*3*3 = 243 > > 3^4 times 3^5 = 81 * 243 = 19,683 (My pocket computer says). > > Now 3^(4+5) = 3^9 OK? > So 3^(4+5) = 3^9 = 3*3*3*3*3*3*3*3*3*3 = 19,683 > > We see that indeed we have that 3^4 times 3^5 = 3^(4 + 5). Both > sides are equal to 19,683. OK? > > 2) verification without computation: > > 3^4 * 3^5 = 3*3*3*3 * 3*3*3*3*3 > = 3*3*3*3*3*3*3*3*3 > = 3^9 > = 3^(4+5) > > OK? > > We use the fact that multiplication is associative a*(b*c) = (a*b)*c > = a*b*c. No need for parenthesis. > > The verification without computation gives an idea how we can > convince ourself of the truth of the general statement: > > a^n times a^m is equal to a^(n+m) > > a^n = a*a*a* ... *a with n occurences of "a". > a^n = a*a*a* ... *a with m occurences of "a". > > a^n times a^m = a^n * a^m = (a*a* .. a) * (a*a*...a) with n > occurences of a in the first parentheses and m occurences of a in > the second parentheses. Of many "a" appears in the right sides: n+m. > > Tell me if this helps. And if and when this satisfies you, you can > read the following (but not before!) > > > ------ > Then I gave a definition: > >> >> This extends to the rationals by defining a^(-n) by 1/a^n. > > I could, like some teacher, just say: accept this as a definition. I > could also provide motivation for such a definition. You have to > separate those two things. > > Accepting the definition, you can already deduce that: > > 5^(-2) = 1/(5^2) = 1/25 = 0.04 > > 2^(-4) = ? (answer: 1/16 = 0.0625) > 4^(-3) = (answer: 1/64 = 0.015625) > 10^(-5) = (answer: 1/100000= 0.00001) > > > Bu why should we accept this definition? Here are the motivation. > > a/b represent a fraction, and its value is the number obtained by > dividing a by b. If a and b are natural numbers, most frequently a/b > will NOT be a natural number. "a/b" is called a fraction. "a" is the > numerator, "b" is the denominator. > The value of fraction are called "rational number". In particular > all natural numbers are rational, but most rational numbers are not > natural number. In term of set, if N = the natural numbers, and Q = > the rational numbers, we have that N is included in Q. (the number 2 > is the value of the fraction 2/1). > > Examples: > > 5/2 = 2.5 (not a natural number!) > 10/5 = 2 (but we were lucky!) > 10/6 = 1.6666... (my pocket computers says). 1.66... is NOT a > natural number. > > Do you agree that (10*a)/(6*a) = 10/6. That is, if you have a > fraction, and if you multiply the numerator by some number, and the > denominator by that same number, you don't change the value of the > fraction. OK? > > So, when we have a fraction, we can multiply or divide the > denominator and numerator by the same number, and this without > changing the value of the fraction: > > So (a*a*a*a)/ (a*a*a) = (by dividing by a both the numerator and the > denominator) (a*a*a)/(a*a) = (the same) (a*a)/a = a. > > Remember: in case of doubt, verify this on simple number. If the > equality (a*a)/a = a seems too much esoteric, verify it with a = > some simple number, like 10 for example. (10*10)/10 = 100/10 = 10, > etc. > > OK up to now? I continue if you tell me if it is OK. But to sum up, > a bit, the definition > > a^(-n) = 1/(a^n) > > is what makes possible to keep the law (a^n * a^m) = a^(n+m) for all > n, no more just positive integers, but on any integers. (where n + (- > m) is defined by n-m). > > You don't have to despair. If you don't understand something, it > means that either you don't know a definition, or you don't recall a > definition, or you don't have integrate a definition (that is, you > are not familiar with the definition, so you can recall it, but > still cannot use it). But I cannot guess where the problem is. I can > guess only that you suffer a lack of practice. > > Just a question to proceed later: do you agree that > > (a*a*a*a)/ (a*a*a) = (a*a*a*a) times 1/(a*a*a) > > with this and what I said, you can guess that: > > 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: > > 10^(n-m)= 10^ -1= 10/1 > > which of course makes no sense at all. Where did I go wrong? > > > > > > Bruno > > > > > > > > > > http://iridia.ulb.ac.be/~marchal/ > > > > > > > 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 -~----------~----~----~----~------~----~------~--~---