August 7th, 2018, 06:06 AM  #1 
Senior Member Joined: Nov 2011 Posts: 222 Thanks: 2  zero in the middle of xaxis
How I prove that zero is between plus infinity to negative infinity?

August 7th, 2018, 06:09 AM  #2 
Senior Member Joined: Oct 2009 Posts: 436 Thanks: 147 
Depends on the specific axioms and definitions you are using. If you list them all, then perhaps we can concoct a proof.

August 7th, 2018, 06:24 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 
Before we try to prove it, perhaps we should determine whether zero is between them or equal to both of them. $\dfrac{1}{\ \infty} = 0 \implies \ \infty = 1 * 0 = 0.$ $\dfrac{1}{\infty} = 0 \implies \infty = 1 * 0 = 0. $ $\therefore \ \infty = 0 = \infty.$ Last edited by JeffM1; August 7th, 2018 at 06:29 AM. 
August 7th, 2018, 08:01 AM  #4 
Senior Member Joined: Jun 2015 From: England Posts: 844 Thanks: 252 
Because zero is finite.

August 7th, 2018, 08:11 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446  
August 7th, 2018, 08:29 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,285 Thanks: 1681  
August 7th, 2018, 08:32 AM  #7 
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446  
August 7th, 2018, 10:00 AM  #8 
Senior Member Joined: Nov 2011 Posts: 222 Thanks: 2  Elaboration text
Why the n^2> infinity greater than the n> infinity? Can one give me an elaboration text on the idea above. (a link) How many infinities there are? Could it explain the post of my above? Last edited by skipjack; August 7th, 2018 at 05:00 PM. 
August 7th, 2018, 12:06 PM  #9 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,852 Thanks: 749 Math Focus: Wibbly wobbly timeywimey stuff. 
In your first post you really shouldn't be using infinity as a number. Let's reframe the question as prove that n < 0 < n when n is very large. The solution I would prefer comes out of Calculus called the intermediate value theorem, but the logic is fairly simple. In order to go from a negative number to a positive number you have to go through 0, don't you? The number system doesn't just "jump" from one real number to another. It is valid to say that $\displaystyle n^2 > n$ for large n. It is not correct to say that $\displaystyle n^2$ represents a "larger infinity" than n. They both represent the same infinity. As to your question: $\displaystyle n^2 > n \implies n^2  n > 0 \implies n(n  1) > 0 \implies n > 0 \text{ or } n > 1$. The case of n > 0 is not strictly true because, by inspection, $\displaystyle n^2 < n$ in this domain. Thus we have for n > 1, $\displaystyle n^2 > n$. Dan Last edited by topsquark; August 7th, 2018 at 12:16 PM. 
August 7th, 2018, 12:34 PM  #10  
Senior Member Joined: Aug 2012 Posts: 1,971 Thanks: 550  Quote:
Consider a line, infinitely long in each direction. Like Euclid's idea of a line. Now pick an entirely arbitrary point on the line and call it 0. Pick another point on the line and call it 1. Now having chosen those two arbitrary points, you can define all the integers, positive and negative whole numbers, based on the unit length between 0 and 1. Then you can define the rationals, all the fractions between the integers. And then you can define the rest of the real numbers by plugging the holes where there are points but not rationals. So you see that the labelling of the points is completely arbitrary. 0 and 1 could be anything you like. Here's another way to look at it. You have a bunch of houses on the street. One day the city council decides to renumber all the houses. The addresses of the houses all change. But the houses themselves remain exactly the same as they were before. A real number is the address of a point on a line. That's all it is. A real number is an address, not the point itself. And addresses are arbitrary, they don't tell you anything about the points themselves. Likewise in a computer program. The address of a variable is not the same as the contents. The address is just the memory location where the value of some variable lives. The address is not the variable. It's only the address.  

Tags 
middle, xaxis 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Capital I with a circle in the middle  calypso  Advanced Statistics  1  December 28th, 2015 01:45 AM 
x axis and y axis Product of slopes is equal to 1  brhum  PreCalculus  4  November 27th, 2014 04:16 AM 
volume xaxis vs yaxis  thesheepdog  Calculus  2  November 10th, 2014 11:23 AM 
how to find the middle of a space?  eftim  Calculus  6  October 5th, 2010 03:05 PM 