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 February 9th, 2016, 06:00 AM #51 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 You have been told, repeatedly, what is wrong with your proof. Your basic difficulty is that you really do not understand "proof by induction". "Proof by induction" shows that a statement is true "for all n". That is, for any positive integer, n. It does NOT prove that a statement is true for "infinite n". February 9th, 2016, 06:32 AM   #52
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Quote:
 Originally Posted by Country Boy You have been told, repeatedly, what is wrong with your proof. Your basic difficulty is that you really do not understand "proof by induction". "Proof by induction" shows that a statement is true "for all n". That is, for any positive integer, n. It does NOT prove that a statement is true for "infinite n".
I said for ALL, ALL, ALL n, whatever name you wish to give that.

Why does everybody keep repeating the same thing over and over again. As I said before, why don't you just assume everybody except me agrees with whatever point it is you are making and you are not going to convince me by saying the same thing over, and over, and over again.

EDIT I was assuming we both were using the same definition of proof by induction. Perhaps I am wrong. What is your definition of proof by induction?

Last edited by zylo; February 9th, 2016 at 06:35 AM. February 9th, 2016, 06:51 AM #53 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Since you won't accept my definition of proof by mathematical induction, how about: "Mathematical induction is a mathematical proof technique, most commonly used to establish a given statement for all natural numbers, although it can be used to prove statements about any well-ordered set. It is a form of direct proof, and it is done in two steps. The first step, known as the base case, is to prove the given statement for the first natural number. The second step, known as the inductive step, is to prove that the given statement for any one natural number implies the given statement for the next natural number. From these two steps, mathematical induction is the rule from which we infer that the given statement is established for all natural numbers." From https://en.wikipedia.org/wiki/Mathematical_induction February 9th, 2016, 07:04 AM #54 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,685 Thanks: 2665 Math Focus: Mainly analysis and algebra And where in that description of induction does it say that the statement is true for an infinite value of $n$? February 9th, 2016, 07:11 AM   #55
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 Originally Posted by v8archie And where in that description of induction does it say that the statement is true for an infinite value of $n$?
It says it's true for all n, ie, all decimals. February 9th, 2016, 07:25 AM #56 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 A decimal representation of a non-negative real number r is an expression in the form of a series, traditionally written as a sum $\displaystyle r=\sum_{i=0}^\infty \frac{a_i}{10^i}$ where a0 is a nonnegative integer, and a1, a2, ... are integers satisfying 0 ≤ ai ≤ 9, called the digits of the decimal representation. The sequence of digits specified may be finite, in which case any further digits ai are assumed to be 0. Some authors forbid decimal representations with a trailing infinite sequence of "9"s. This restriction still allows a decimal representation for each non-negative real number, but additionally makes such a representation unique. The number defined by a decimal representation is often written more briefly as $\displaystyle r=a_0.a_1 a_2 a_3\dots.\,$ That is to say, a0 is the integer part of r, not necessarily between 0 and 9, and a1, a2, a3, ... are the digits forming the fractional part of r. Both notations above are, by definition, the following limit of a sequence: $\displaystyle r=\lim_{n\to \infty} \sum_{i=0}^n \frac{a_i}{10^i}.$ from: https://en.wikipedia.org/wiki/Decima...epresentations -------------------------------------------------------- $\displaystyle r=.a_1 a_2 a_3\dots.\,$ 1,2,3,4...... February 9th, 2016, 07:36 AM   #57
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Quote:
 Originally Posted by zylo It says it's true for all n, ie, all decimals.
No it doesn't. It says "established for all natural numbers".

So now, to get an infinite $n$, you have to identify an infinite natural number. Can you do that? February 9th, 2016, 07:41 AM   #58
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 Originally Posted by zylo A decimal representation of a non-negative real number...
Yes. I don't think anyone here is struggling to understand what a decimal representation of a number is. What's your point? February 9th, 2016, 08:30 AM #59 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 pi can be expressed to any number n decimal places. There is no value of n=$\displaystyle \infty$ that gives pi. All you have to do is prove you can express pi for any number (all) decimal places, ie, the decimals are countable, ie the reals are countable. February 9th, 2016, 08:35 AM   #60
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 Originally Posted by zylo pi can be expressed to any number n decimal places.
No. $\pi$ can only be approximated to $n$ decimal places. Tags counting, irrational, numbers ,

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### Can you count to an irrational number

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