Re: Theorem on Natural Numbers

MoeBlee wrote:
On Nov 14, 12:48 pm, rabbits77 <rabbit...@xxxxxxxxxxx> wrote:
MoeBlee wrote:
On Nov 14, 11:02 am, rabbits77 <rabbit...@xxxxxxxxxxx> wrote:
Now assume a^k divides b^k => a divide b.
Show that this works for a^(k+1) divides b^(k+1) too and you are done!
How do you show it?
From basis step we know that a divides b

The basis step is a^1|b^1 -> a|b. That is not a problem.

Then the inductive step is that for all a and b, we have a^k|b^k -> a|
b. Then we asssume a^(k+1)|b^(k+1). And the mission then is to show a}
b.

We assume a^k divides b^k.

No, we don't. We assume a^(k+1)|b^(k+1) with the mission then of
showing a|b upon the inductive assumption that for all a and b, we
have a^k|b^k -> a|b.

b^(k+1)/a^(k+1)= (b)...(b) . For k+1 many a's and b's
---------
(a)...(a)
which is k+1 many repetitions of the basis step a divides b.
Or, in other words, (b^k/a^k)*(b/a)...the product of two statements that
are true.

That is not any form of mathematical induction with which I am
familiar. Here is what we do have:

(1) Assume, for all a and b, we have a^k|b^k -> a|b.

(2) Assume a^(k+1)|b^(k+1).

(3) Show: a|b.

huh? Assume a^(k+1)|b^(k+1) ?
Enjoy this link. It covers basic induction. In particular look at
http://tinyurl.com/6sym8z
In particular look at the very beginning of pg 2. This is exactly what I did.

There is no "repetitions of basis step" schema of induction I am
familiar with. Looks like handwaving to me. Otherwise, please refer me
to the exact induction schema you have in mind.

You have the link above.
"repetitions of basis step" is a straightforward
english description of pointing something out. Not formal mathematics.
Your english may not be good enough for this group. Consider a math news
group in your native language.
.

Relevant Pages

  • Re: Theorem on Natural Numbers
    ... Show that this works for a^divides b^too and you are done! ...  From basis step we know that a divides b ... There is no "repetitions of basis step" schema of induction I am ... to the exact induction schema you have in mind. ...
    (sci.math)
  • Re: Theorem on Natural Numbers
    ... From basis step we know that a divides b ... That is not any form of mathematical induction with which I am ... As with any implication, in order to prove that A-->B, we may assume ...
    (sci.math)
  • Re: Theorem on Natural Numbers
    ... form of weak induction, since weak induction does not have a clause ... "repetitions of basis step". ... He then argued that if a^k divides b^k, ...
    (sci.math)
  • Re: divisibility problem
    ... I have that okay. ... Would you show me more overtly how you set up that induction? ... divides n!. ... The tax collector collects the tax in a strange way. ...
    (sci.math)
  • Re: Help w/ Log335 homework
    ... My instructor tried to give us hints, but none of them made sense to ... This doesn't necessarily need induction. ... q, and 2 divides q?) ... That should be more than enough of a hint. ...
    (sci.math)