My Math Forum Counting the Irrational Numbers

 Topology Topology Math Forum

 February 2nd, 2016, 10:03 AM #11 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,197 Thanks: 898 Math Focus: Wibbly wobbly timey-wimey stuff. I'm just being curious here... zylo: Have you looked up a proof that claims the real numbers are uncountable? What's so wrong with the Cantor proof? -Dan
 February 2nd, 2016, 11:38 AM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra A decimal with $m \in \mathbb N$ decimal places terminates after $m$ decimal places (otherwise it doesn't have $m$ decimal places). An infinite decimal does not terminate, by definition. So you claim to have decimals that do not terminate that also terminate after $m$ decimal places for some $m \in \mathbb N$. How does that work? You even agreed the other day that all natural numbers are finite. Last edited by skipjack; February 2nd, 2016 at 12:34 PM.
February 2nd, 2016, 02:54 PM   #13
Math Team

Joined: Nov 2014
From: Australia

Posts: 689
Thanks: 244

Quote:
 Originally Posted by zylo If you can count n decimal places you can count n+1 decimal places, therefore by induction you can count ALL (countably infinite) decimal places and therefore all real numbers.
This is a common misconception about mathematical induction. You can only conclude the result for all finite $n$. Induction gives no information about the infinite case.

February 2nd, 2016, 03:16 PM   #14
Global Moderator

Joined: Dec 2006

Posts: 20,746
Thanks: 2133

Quote:
 Originally Posted by zylo If you can count n decimal places you can count n+1 decimal places, therefore by induction you can count ALL (countably infinite) decimal places and therefore all real numbers. If you don't accept induction, that is another matter.
You used induction incorrectly. In a proof by induction, it's not legitimate to change the definition of the proposition.

You weren't originally counting the number of decimal places; you were considering the decimals of length $n$, where $n$ is a whole number. Your proposition was that these are countable, which was proved by induction to be correct. However, no such real is irrational, so you didn't prove anything about the irrational reals. In claiming to have done so, you effectively changed the definition of the proposition you were considering.

February 4th, 2016, 05:20 PM   #15
Banned Camp

Joined: Mar 2015
From: New Jersey

Posts: 1,720
Thanks: 125

Quote:
 Originally Posted by skipjack You used induction incorrectly. In a proof by induction, it's not legitimate to change the definition of the proposition. You weren't originally counting the number of decimal places; you were considering the decimals of length $n$, where $n$ is a whole number. Your proposition was that these are countable, which was proved by induction to be correct. However, no such real is irrational, so you didn't prove anything about the irrational reals. In claiming to have done so, you effectively changed the definition of the proposition you were considering.
P(n): The number of n-place decimals is countable.
p(n) -> p(n+!): The number of n+1-place decimals is countable.
P(n) true for all n (to infinity): All decimals are countable, ie, the reals are countable.

I didn't change the proposition

Last edited by skipjack; February 4th, 2016 at 05:47 PM.

 February 4th, 2016, 05:35 PM #16 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra Yet again: THERE ARE NO INFINITE NATURAL NUMBERS SO YOUR PROOF DOES NOT REFER TO ANY INFINITE DECIMALS ONLY FINITE ONES No proof by induction on $n$ proves anything for infinite $n$, because proofs by induction use the natural numbers, which are all finite. Repeating your error over and over again will never correct it. It just shows how little you understand about what you are talking about.
 February 4th, 2016, 06:58 PM #17 Global Moderator   Joined: Dec 2006 Posts: 20,746 Thanks: 2133 A proof by induction of a proposition P($n$) must establish that P($n$) holds for all finite $n$ because there is no such thing as an infinite value of $n$ (every integer is finite). Such a proof can deal with infinitely many values of $n$, but not any infinite value of $n$, because no such value exists. Although the word "countable" was used in the proof, the number of $n$-place decimals for a given value of $n$ is finite, but increases without bound as $n$ increases indefinitely (the sequence obviously couldn't have a finite upper bound). Hence the proof establishes that the set of all terminating decimals (by definition, every $n$-place decimal terminates at its $n$th place) is infinite, but countable. The reference to "all decimals", as distinct from all terminating decimals, effectively changed the proposition to include non-terminating decimals that hadn't previously been considered in the proof. Do you now understand, zylo, that you changed the proposition? If not, which of my above five sentences do you fail to understand?
 February 4th, 2016, 07:12 PM #18 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Induction applies for all n from 1 to infinity. I acknowledge that you don't accept that. "Mathematical induction is a mathematical proof technique, most commonly used to establish a given statement* for all natural numbers" quoted from (* added): https://en.wikipedia.org/wiki/Mathematical_induction *The n place decimals are countable. What do you think an n-place decimal is? Last edited by zylo; February 4th, 2016 at 07:15 PM. Reason: reference added
February 4th, 2016, 07:25 PM   #19
Math Team

Joined: Nov 2014
From: Australia

Posts: 689
Thanks: 244

Quote:
 Originally Posted by zylo What do you think an n-place decimal is?
A number with $n$ decimal places, for $n\in\mathbb{N}$. Also known as a rational number.

What's that got to do with your proof that the irrationals are countable?

February 4th, 2016, 07:35 PM   #20
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,663
Thanks: 2642

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by zylo Induction applies for all n from 1 to infinity.
There is no number "infinity". You obviously don't know what the mathematical meaning of the phrase "to infinity" is. Induction is a method of proving a propositions for the natural numbers. All natural numbers are finite. "Infinity" is not a number and it is not a natural number. A proof by induction proves a proposition for finite $n$.

An $n$-place decimal is one that terminates after $n$ decimal places. In infinite decimal does not terminate at all.

 Tags counting, irrational, numbers

,

,

### Can you count to an irrational number

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post AspiringPhysicist Abstract Algebra 1 October 12th, 2014 04:15 AM Albert.Teng Algebra 4 February 12th, 2014 04:55 PM niki500 Number Theory 5 October 7th, 2012 09:10 PM Deb_D Number Theory 12 February 20th, 2011 11:50 AM Mighty Mouse Jr Algebra 1 October 16th, 2010 07:46 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top