Definitions of 0 and infinity
- Thread starter K.I.L.E.R
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That's what I thought but from past experience things like 0 and infinity are undefined.
Certainly powers are recursive functions and looking at 1^ifty from that point of view it is undefined as it would never be complete however 1^x is just going to be 1 anyway.
Certainly powers are recursive functions and looking at 1^ifty from that point of view it is undefined as it would never be complete however 1^x is just going to be 1 anyway.
Their are different types of infinity. If you replace infinity with 1/0 for example then those identities make sense. The general practice is when infinity crops up as an answer you should not reduce to infinity, because after later operations the quantity may become finite again. You can see how 0/0, 1/0, and 2/0 are three different infinite values. If (1/0 = infinity) does (infinity* 0 = 1) it does not, but (1/0 *0 =1) is true.
It's similiar to how irrational numbers have infintie decimal places yet all irrational numbers are not equal. If we think about the irrational numbers as a fraction then the numerator would be one infinite number and the denomator a different one.
It's similiar to how irrational numbers have infintie decimal places yet all irrational numbers are not equal. If we think about the irrational numbers as a fraction then the numerator would be one infinite number and the denomator a different one.
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I would say that is undefined.
Since 1^N == 1, then lim 1^N as N->inf is still 1. I'd argue that it might be well defined.
Of course, you can't do Xk^N where Xk->1 but that surely is another matter entirely.
No, 1/0 * 0 = 1 is not true. 1/0 * 0 is undefined.
Let x = 1/0 * 0, x will represent the value of this equation. If you think that a 0 in the denominator can be "cancelled", multiply both sides by 0.
0 * x = 1 * 0
There are an infinite number of values X that satisfy this equation.
0 has no multiplicative inverse, period.
Irrationals cannot be defined solely by a ratio of two integers, even ones with an infinite number of digits.
Attempting to divide by zero, even if you don't explicitly do it, is what causes this to be false.
a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1
a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1
Unlike something like R/infinity = 0 where R is real number, infinity^0 is indeterminate.
For example: lim (x^(1/x)) when x-->infinity
lim {x^(1/x)} when x-->infinity
= lim {e^ln[x^(1/x)]} when x-->infinity
= lim {e^[(ln x)/x]} when x-->infinity
= lim {e^[(x-1)/x]} when x-->infinity
= lim {e^[1-1/x]} when x-->infinity
= e
See the limit doesn't equal to 1.
Blazkowicz
Legend
even 0^0 is undefined. that tells me not to mess with infinity^0 (I have a proof but it doesn't fit in the forum's margin)
