My Math Forum When do I know that my proof is good enough?

 September 10th, 2013, 04:51 AM #1 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 When do I know that my proof is good enough? Hello there, I am a first year university student of physics. I was asked to prove that any repeating decimal numbers such as 0.333333333333333... are always rational. I will try to make this proof, but how do I know that a proof is good enough? Thank you for your time. Kind regards, Marius
 September 10th, 2013, 04:57 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: When do I know that my proof is good enough? You need a proof first. Then, check it yourself and, if you're still not sure, get someone else to check it.
September 10th, 2013, 05:10 AM   #3
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Re: When do I know that my proof is good enough?

Quote:
 Originally Posted by Pero You need a proof first. Then, check it yourself and, if you're still not sure, get someone else to check it.
Thank you for your reply. I do not have any formal training in writing proofs, they have been shown to me many times, but I need some sort of outline of what is OK in a proof, and what is not OK.

Where can I learn what is allowed in a proof?

Kind regards,
Marius

 September 10th, 2013, 05:18 AM #4 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: When do I know that my proof is good enough? In my proof, is it okay to state that; If the decimals are repeating, it is because there is always a remainder when two integers are divided? Thereby completing the proof? Thank you for your time. Kind regards, Marius
 September 10th, 2013, 05:29 AM #5 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: When do I know that my proof is good enough? That's not a proof of anything. Here's a hint for this proof (or any proof). 1) Try to think why it is true. 2) Try to think why it may not true. If you are trying to prove something, it can be very useful to try to find a counterexample. Trying to disprove it will often give you vital clues as to why it is true. In this particular case, you are going to have to tackle to definition of 0.33333... You're not going to find a proof without that. I'd start by proving that 0.3333... is rational. Then, maybe do another one 0.444... Then, maybe you can see a pattern of how to prove it it general.
 September 10th, 2013, 05:31 AM #6 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: When do I know that my proof is good enough? Actually, before you try to prove this, you should first say what 0.333... actually means. And, you can't say "the 3's go on for ever" as that is not precise enough. The first step, therefore, is to define 0.333...
 September 10th, 2013, 05:51 AM #7 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: When do I know that my proof is good enough? Woah, that sounds like a fantastic challenge. I will get to it right away! Kind regards, Marius
 September 10th, 2013, 09:44 AM #8 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: When do I know that my proof is good enough? Thank you. I found a way to convert repeating decimal numbers to fractions with integers in the numerator and denomerator, which by definition are rational. This is how: A repeating decimal number, contains a finite number of digits which repeat in sequence. For example, the number n contains digits on this form: $d_{i}$ $n= d_0,d_1d_2d_3d_4...d_m \bar{\, d_1d_2d_3d_4...d_m}$. They can be composed in any way you like, but the digits have to be repeating. $n \cdot 10^m= d_0d_1d_2d_3d_4...d_m, \bar{\, d_1d_2d_3d_4...d_m}$ $(n \cdot 10^m) - n= n( 10^m - 1) \in \mathbb{Z}$ $( 10^m - 1) \in \mathbb{Z}$ $n= \frac{n( 10^m - 1)}{( 10^m - 1)}$ $n( 10^m - 1)$ and $( 10^m - 1)$ are obviously integers, and therefore n is by definition rational. ? How is that? Thank you for your time. Kind regards, Marius
 September 11th, 2013, 08:56 PM #9 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: When do I know that my proof is good enough? Looks fine to me!
 September 15th, 2013, 11:42 PM #10 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: When do I know that my proof is good enough? This is not quite right, as you've used n to be both the fraction and the whole number that represents the repeating decimal. You need to have something like: $n= 0.d_1d_2d_3d_4...d_md_1d_2d_3d_4...d_m...$ $d= d_1d_2d_3d_4...d_m$ Then: $n \cdot 10^m= d_1d_2d_3d_4...d_m.d_1d_2d_3d_4...d_m...$ $(n \cdot 10^m) - n= d$ $n(10^m - 1)= d$ $n= \frac{d}{(10^m - 1)}$ $d$ and $( 10^m - 1)$ are obviously integers, and therefore n is rational. The other way to derive this is to note that a repeating decimal is an infinite geometric progression with first term: $0.d_1d_2d_3d_4...d_m$ And common ratio $10^{-m}$ The sum of this progression is then: $n= \frac{0.d_1d_2d_3d_4...d_m}{1-10^{-m}}$ And, multiplying through by $10^m$ gives: $n= \frac{d}{10^m - 1}$

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