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 June 25th, 2010, 07:00 AM #1 Newbie   Joined: Jun 2010 Posts: 2 Thanks: 0 Infinity... Taking infinity to be a number. Then since the product of the gradients of two perpendicular lines equals -1 (i.e. m_1 x m_2=-1) then this implies that, ? x 0=-1 , since a vertical line with an infinite gradient is perpendicular to one with a gradient of 0. Can someone clear this up for me....
 June 25th, 2010, 07:20 AM #2 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: Infinity... The mistake in your proof is the first sentence.
 June 25th, 2010, 07:25 AM #3 Newbie   Joined: Jun 2010 Posts: 2 Thanks: 0 Re: Infinity... is this just another proof why you cannot take it to be a number then? i understand that it is a concept, but the more i research the more i find people attempting arithmetic operations using infinity!!
 June 25th, 2010, 08:22 AM #4 Newbie   Joined: Jun 2010 Posts: 14 Thanks: 0 Re: Infinity... 0 * infinity is of the indeterminate form. It is up there with such things as 0/0, infinity/infinity, and infinity - infinity. The best way to think of indeterminate form is things who's answers depend on which rules you use. In the case of 0/0, we know that 0/anything = 0, anything/0 = infinite, and anything/itself = 1. The issue with 0 * infinity is that 0 * anything = 0 and infinity * anything = infinite. The cool thing about indeterminate forms is that depending on how we think of them, we can essentially make them equal to anything we want, including -1.
 June 25th, 2010, 09:20 AM #5 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Infinity... This, being false, could be said as well, if you´d define infinity as a number. There are so many numbers between 0 and 1 as between 1 and $\infty$, because every number between 0 and 1, m, is $\frac{1}{m}$ which than is between 1 and $\infty$. Examples: $\frac{1}{0.5}=2$, $\frac{1}{4}=0.25$. How is this properly expressed? Anyway, Their are so many numbers between 0 and 1 as between 1 and 2. every number, n, between 0 and 1, has a solution between 1 and 2. Examples: 0.5+1=1.5, 0.25+1=1.25 So there would be as many numbers beween 1 and 2 as between 1 and $\infty$... Hoempa
 June 25th, 2010, 10:49 AM #6 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: Infinity... I can argue that infinity is as much of a number as zero is.
June 25th, 2010, 10:51 AM   #7
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Re: Infinity...

Quote:
 I can argue that infinity is as much of a number as zero is.
Go for it...

Infinity is a concept, not a number. Tell me the value of infinity and I'll give you a number larger.

June 25th, 2010, 11:06 AM   #8
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Re: Infinity...

Quote:
 Originally Posted by Hoempa This, being false, could be said as well, if you´d define infinity as a number. There are so many numbers between 0 and 1 as between 1 and $\infty$, because every number between 0 and 1, m, is $\frac{1}{m}$ which than is between 1 and $\infty$. Examples: $\frac{1}{0.5}=2$, $\frac{1}{4}=0.25$. How is this properly expressed? Anyway, Their are so many numbers between 0 and 1 as between 1 and 2. every number, n, between 0 and 1, has a solution between 1 and 2. Examples: 0.5+1=1.5, 0.25+1=1.25 So there would be as many numbers beween 1 and 2 as between 1 and $\infty$... Hoempa
This argument is not correct though, as it fails to include irrational numbers. Irrationals, remember, are not of the form p/q. As it turns out, there are many more irrational numbers between 1 and $\infty$ than there are irrational numbers between 0 and 1.

 June 25th, 2010, 11:07 AM #9 Senior Member   Joined: Apr 2010 Posts: 215 Thanks: 0 Re: Infinity... There are numbers larger than infinity, but those, (just like infinity) aren't real. There's no real number larger than my infinity concept. That is the key part. Some similarities between zero and infinity: Zero is neither positive nor negative. Same thing goes for infinity. For any x: $x\cdot0= 0$ $x\cdot\infty= \infty$
June 25th, 2010, 11:39 AM   #10
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Re: Infinity...

$\sqrt{2}$ is irrational. $1<\sqrt{2}<\infty$
$\frac{1}{\sqrt{2}}=0.5\sqrt{2}$. $0<0.5\sqrt{2}<1$
Is this false or do you mean something else?

As far as I know, Infinity, if defined as a number, will have infinite values, not 1 value, like 2 has. $\infty+1=\infty$

Quote:
 Originally Posted by brangelito For any x: $x\cdot 0=0 \\ x\cdot \infty=\infty$
What if $x=\infty$?
In your argument, you could substitute $\infty$, since it´s a number. Now we (I) are (am) referring to the OP´s post.
See OP´s post:
He/she argues:
$\infty \cdot 0=-1$

$\infty$ is positive.
$\lim_{x \rightarrow \infty}$ says basically you are substituting a very large (largest?), undefined number.

Hoempa

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