SeaCalMaster said:
Lukeje said:
BloodSquirrel said:
Lukeje said:
One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.
That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.
I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.
No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.
Lesson time!
When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0. As for whether or not infinity + n = infinity, that depends on the exact definition of addition you're using. However, a definition that makes it true that infinity + n = infinity for all n breaks associativity.
Lesson time!
In number systems where infinity is a member, infinity is almost always considered to not have an additive inverse.
Just like 0 is an exception and doesn't have a multiplicative inverse in the Real numbers.
Hence, just as 0/0 zero is undefined, infinity - infinity is undefined.
The first doesn't stop 0 from being a useful number, nor does the latter stop infinity from it.
Maze1125 said:
SeaCalMaster said:
Maze1125 said:
Lukeje said:
TheTechnomancer said:
Lukeje said:
TheTechnomancer said:
Well i'm not sure but my maths text book says 1/0 is infinity so no ofense but i'll trust that over you.
Are you sure it doesn't just say that the limit of 1/
x as
x->0 is infinity? Infinity is a tricky concept...
Pretty sure as some questions in the book required you to use 'infinity = 1/0' in order to get to the correct answer.
Example please?
Riemann Sphere [http://en.wikipedia.org/wiki/Riemann_sphere]
AAH WHY DO THE ANALYSTS HAVE TO RUIN EVERYTHING
THE EXTENDED COMPLEX NUMBERS AREN'T EVEN A GROUP UNDER ADDITION
Also, it's not technically correct to say that the limit of 1/
x as
x->0 is infinity. The series diverges, so it's not even really true that the limit exists.
If you're using a set where -infinity = infinity (such as the Riemann Sphere), then 1/x in fact converges as x->0, as the limits from both sides are equal.
No, it doesn't, at least not under the standard metric. If we use the standard definition where |infinity - n| = infinity (if n != infinity), then the distance from 1/x to infinity for any (non-zero) value of x is infinite.
So you're claiming that f(x) = x -/> infinity as x -> infinity?
Pro-tip, the definition of tending to infinity is different to the definition of tending to a finite number.
1/x -> infinity as x -> 0 from above, and 1/x -> -infinity as x -> 0 from below.
So, if infinity = -infinity, the limits from both sides are the same, and so, by the definition of convergence, the function converges at 0.