My Math Forum The Biggest Number

 Algebra Pre-Algebra and Basic Algebra Math Forum

November 22nd, 2017, 11:05 PM   #11
Senior Member

Joined: Feb 2016
From: Australia

Posts: 1,591
Thanks: 546

Math Focus: Yet to find out.
Quote:
 Originally Posted by shaharhada But I don't understand a thing in the last post. Please, someone will write it in a simple English. Thank you!!! You are welcome!!!! [If you answer to this post and to my request.]
It's a proof showing that there is no 'largest number'. See countryboy's post #2 for an explanation in simple English.

 November 23rd, 2017, 03:06 AM #12 Senior Member   Joined: Nov 2011 Posts: 194 Thanks: 2 What about this idea I think iff A is the biggest number, there is also A plus 1 [A+1] number. So there no biggest number in real number system. But in fractions smaller than one, one is the biggest. Right? {Also, if it is not inlcude in the this number system because 0.9999999... = 1]
 November 23rd, 2017, 03:42 AM #13 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra No. There is no biggest fraction smaller than 1. $0.999\dots$ and $1$ just represent two infinite sequences that have the same integer limit. The limit is the real number. This is a definition of the real numbers. Decimal notations aren't numbers, they are representations of numbers.
 November 23rd, 2017, 03:44 AM #14 Senior Member   Joined: Nov 2011 Posts: 194 Thanks: 2 Why? You have an example that I am not right. Why I can't define 0.999... as 1.
 November 23rd, 2017, 03:48 AM #15 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 No, there is no largest "fraction less than one". Suppose there were - that is, suppose $\displaystyle 0< \frac{a}{b}< 1$. We can, of course, take both a and b positive. Now look at $\displaystyle \frac{\frac{a}{b}+ 1}{2}$, the "average" of the fraction and one. $\displaystyle \frac{\frac{a}{b}+ 1}{2}= \frac{a+ b}{2b}$. We can prove that is "a fraction less than one". It is a fraction and $\displaystyle 1- \frac{a+ b}{2b}= \frac{2b- (a+ b)}{2b}= \frac{b- a}{2b}$. $\displaystyle b > a$ because $\displaystyle \frac{a}{b}< 1$ so this is positive so 1 is larger than $\displaystyle \frac{a+ b}{2b}$. We can prove this fraction is larger than $\displaystyle \frac{a}{b}$. $\displaystyle \frac{a+ b}{2b}- \frac{a}{b}= \frac{a+ b- 2a}{2b}= \frac{b- a}{2b}$. Again, that is positive as $\displaystyle \frac{a+ b}{2b}$ is larger than $\displaystyle \frac{a}{b}$. That is, for any fraction less than 1 there exist another, larger, fraction less than 1. You don't need to "define 0.999... as 1", it is 1! But because it is equal to 1, it is not "a fraction less than 1" so it is not the "largest fraction less than 1". Last edited by skipjack; November 23rd, 2017 at 04:13 AM.
 November 23rd, 2017, 03:56 AM #16 Senior Member   Joined: Nov 2011 Posts: 194 Thanks: 2 What shall I do I need to put on (a/b) the fraction 0.999... or 1, to see that the expressions are equal or they are not equal and where it is going to. Thank you for consider my message. Last edited by shaharhada; November 23rd, 2017 at 03:59 AM.
November 23rd, 2017, 04:03 AM   #17
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,305
Thanks: 2443

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by shaharhada Why? Why I can't define 0.999... as 1.
0.999... is equal to 1, but 0.999... isn't a fraction less than one (because it is one).

As to "why can't I define...": you can define anything you like, but if it doesn't agree with the standard system, it isn't the standard system. You can study your system to see if it's internally consistent and whether it has any interesting or useful results, but any such results are about your system, not the standard one.

 November 23rd, 2017, 04:13 AM #18 Senior Member   Joined: Nov 2011 Posts: 194 Thanks: 2 Thanks... I get it. Thank you!!!
 November 23rd, 2017, 05:37 AM #19 Senior Member   Joined: May 2016 From: USA Posts: 1,029 Thanks: 420 The short and simple answer is that there is no largest real number. If L is the largest number, then L + 1 is larger than the largest number, which makes no sense at all.
 November 24th, 2017, 04:03 AM #20 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 Which was said 18 responses ago. Thanks from Joppy

 Tags biggest, number

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post shaharhada Algebra 8 December 30th, 2017 06:04 PM ungeheuer Computer Science 0 April 13th, 2014 05:05 AM Telo Algebra 1 July 31st, 2013 11:04 AM JC Applied Math 0 January 16th, 2012 02:20 PM deanmullen10 Number Theory 11 May 31st, 2010 10:25 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top