Infinity subtracted from infinity is undefined.

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In summary, the conversation discusses the concept of infinity and how it relates to arithmetic operations. It is pointed out that infinity is not a "real" number and cannot be used in arithmetic calculations. The conversation also mentions that there are different ways to extend the number systems to include infinity and that operations involving infinity may or may not make sense depending on the system being used. Additionally, the conversation delves into the use of l'Hopital's rule and alternate definitions and methods to deal with expressions involving infinity in certain theories. Overall, the conversation highlights the complexities and nuances surrounding the concept of infinity in mathematics and physics.
  • #36
sihag said:
i came across lim x.logx, x --> 0
as equal to 0.
The logx component part would go to - infinity, and that multiplied with a quantity tending to 0 is zero. Does that mean 0 times any quantity (infinite or finite) is always 0?
Before I answer your question, please answer mine: why would you think that might mean 0 times an infinite quantity is always 0?
 
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  • #37
I am getting mightily tired of these "infinity" -threads and what can, or cannot do with them, and that it is somehow strange that some operations remain undefined.

Let us look at the number system consisting of the numbers:

0, 1 and "many"

We let 0 be distinct from 1, and we have relations like 1+1=many, many+1=many and so on.

Numerous undefined relations will naturally occur within such a number system.
 
  • #38
sihag said:
i came across lim x.logx, x --> 0
as equal to 0.
The logx component part would go to - infinity, and that multiplied with a quantity tending to 0 is zero. Does that mean 0 times any quantity (infinite or finite) is always 0?

Slightly off topic, but what's the proof for the sum of two irrational numbers to be irrational?
Sometimes it helps to check if there are trivial counterexamples to what you've claimed. For instance, 1/x^2 tends to infinity as x->0 from the right. x tends to 0 as x->0 from the right. Does the product tend to zero?

For the second question, what is sqrt(2)+(-sqrt(2))? Is it irrational?
 
  • #39
I meant besides the trivial case, for irrational sums.

I stand corrected on the 0 times arbitrarily large quantity case.
Thank you.
 
  • #40
How about (pi + e + 6) + (-pi - e)? If you want to exclude that case as well, it's going to be impossible to define what "non-trivial" means. I think what you want is that an irrational plus a rational is irrational, which is easy to prove.
 
  • #41
Well, it might not be impossible for him to define what he means by "the trivial cases " in a mathematically meaningful way.

Until he does, though, we can't judge whether the sum of two irrationals in his "non-trivial" cases must equal an irrational.
 
  • #42
The only way to define "non-trivial" so that his statement is true is that the sum of the two numbers is not a rational, but then it's vacuously true.
 
  • #43
zhentil said:
The only way to define "non-trivial" so that his statement is true is that the sum of the two numbers is not a rational, but then it's vacuously true.

Not too sure about that.
It might be that we can separate irrationals into two fairly large classes (by some interesting criterion) and that surprisingly, this criterion would imply, throuygh lengthy proofs, that the sum of two such irrationals necessarily would be irrational.

This could well an important result, but we shouldn't hope for, or expect, its materialization.
 
  • #44
Could you be more specific? I have a feeling that's still circular.
 
  • #45
I thought we didn't know if (pi + e) was rational or irrational.
 
  • #46
by non-trivial i meant 'not using the negative inverse of an irrational to cancel it out in the expression'.
For example sqrt 2 + sqrt 3 is a non trivial case.
How does one prove that
a + b = irrational, whenever a not equal to b, a,b both being irrational.
 
  • #47
sihag said:
by non-trivial i meant 'not using the negative inverse of an irrational to cancel it out in the expression'.
The irrational number (1 - sqrt 2) is not the additive inverse of sqrt 2. Nor is
[tex]\sqrt{3 - 2 \sqrt{2}}.[/tex]
 

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