November 22nd, 2017, 11:05 PM  #11 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,610 Thanks: 550 Math Focus: Yet to find out.  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: 222 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: 7,342 Thanks: 2465 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: 222 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: 3,261 Thanks: 894 
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: 222 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: 7,342 Thanks: 2465 Math Focus: Mainly analysis and algebra  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: 222 Thanks: 2  Thanks...
I get it. Thank you!!! 
November 23rd, 2017, 05:37 AM  #19 
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 
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: 3,261 Thanks: 894 
Which was said 18 responses ago.


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