November 10th, 2017, 04:02 AM  #1 
Senior Member Joined: Nov 2011 Posts: 247 Thanks: 3  The Biggest Number
Is there a notation for the Biggest Number that exist? If not, why?

November 10th, 2017, 04:07 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
No, because there is NO "biggest number that exist". Given anything that purported to be the "biggest number" adding 1 gives an even bigger.

November 10th, 2017, 04:09 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 332 Thanks: 42 
The biggest number is $\displaystyle B=\underbrace{99999....9}_{\infty}$ Funny 
November 10th, 2017, 04:27 AM  #4 
Newbie Joined: Oct 2017 From: Italy Posts: 15 Thanks: 0 
Yes it exist in $\overline{\mathbb{R}}$, it's $+\infty$ 
November 10th, 2017, 04:27 AM  #5 
Senior Member Joined: Jun 2015 From: England Posts: 891 Thanks: 269 
Now you know that there is no largest number, ask yourself this is 100 larger than or smaller than +10? 
November 10th, 2017, 04:40 AM  #6 
Senior Member Joined: Nov 2011 Posts: 247 Thanks: 3  Notation
What is the notation of R with line above it? Thank for answering. I still not understood "Why there isn't a notation for the biggest number that exist"? So, If it didn't exist, is there a proof not by adding one to the biggest number that I found but another proof. I want to see another proof. Thank you If You give an ANOTHER Example.... 
November 22nd, 2017, 06:23 PM  #7 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 630 Thanks: 85  
November 22nd, 2017, 06:34 PM  #8 
Senior Member Joined: Aug 2012 Posts: 2,136 Thanks: 622  I think the Rbar notation is for the extended real number system, which includes symbols for positive and negative infinity. I don't think that's the standard notation. Perhaps that poster means the topological closure of the reals. In that case the intention is clear but the notation's inaccurate or at least nonstandard.

November 22nd, 2017, 06:39 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 Math Focus: Mainly analysis and algebra 
I think he's referring to $\bar{ \mathbb R}$ which is the extended real numbers. We know that $2$ exists and that $2 > 1$, so if a largest number $L$ exists, it must satisfy $L \ge 2$. Thus $L^2$, which exists because the reals are closed under multiplication, satisfies $L^2 \ge 2L > L$. This contradicts the assumption that $L$ is the largest number. $\square$ 
November 22nd, 2017, 11:49 PM  #10 
Senior Member Joined: Nov 2011 Posts: 247 Thanks: 3  Sorry,`
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.] 

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