Let A be the set of positive integers, i.e. A = {1, 2, 3, ...}. Let B be the set of non-negative integers, i.e. B = {0, 1, 2, 3, ...}
Let n be the value of any arbitrarily selected member from A. Then, n can be matched up one-for-one with the corresponding member in B that has value (n - 1).
So, then, if we say set A has ∞ members, then logically B has ∞ + 1 members, the additional member being zero. However, the fact that a one-for-one relationship can be demonstrated must mean that ∞ and ∞ + 1 are equivalent.
2007-06-29 04:04:38
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answer #1
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answered by Anonymous
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it somewhat is real that lim n->? [a million^n] = a million. I certainly won't argue with that. to place it yet differently: in case you regard the backside as *fixed* on the quantity a million, and additionally you enable the exponent flow off to ?, then the shrink is a million. --- the rationalization that a million^? is seen an indeterminate type is that if, as a replace of with regards to a million as fixed, you have the backside *coming near* a million, and the exponent coming near ?, then there is not adequate practise to be certain the shrink. alwbsok has already given an extremely sturdy occasion of why: The expression (a million + a million/n)^n. the backside of this expression has a tendency to a million as n has a tendency to ?, and the exponent has a tendency to ? as n has a tendency to ?, however the entire expression (a million + a million/n)^n has a tendency to e, not a million: (a million + a million/a million)^a million = 2 (a million + a million/10)^10 ? 2.594 (a million + a million/a hundred)^a hundred ? 2.705 (a million + a million/one thousand)^one thousand ? 2.717 etc. it somewhat is a risk to coach (making use of L'well being center's Rule) that lim n->? [(a million + a million/n)^n] = e. The expression a million^? is named an "indeterminate type" in the desire of combating you from reasoning interior of here (incorrect) way: "To compute lim n->? [(a million + a million/n)^n], we first see that a million + a million/n has a tendency to a million and n has a tendency to ?, so the shrink would desire to be a million^?, that's a million." the undertaking with this (incorrect) reasoning is that, only because of the fact (a million + a million/n) is on the factor of a million and n is on the factor of ?, it *does not save on with* that (a million+ a million/n)^n is on the factor of a million. --- Your textbook (probable) does not explicitly state that a million^? is undefined--announcing it somewhat is an "indeterminate type" only ability what I stated above: in case you have an expression of the type f(x)^[g(x)], and additionally you attempt to be certain its shrink, and once you're taking the shrink of f(x) and g(x) independently, you get something of the type a million^?, then this is not a risk to end something from that by myself. --- In some experience, defining a million^? to be a million is the main existence like definition a risk (and a few ingredients do say a million^? = a million), yet different ingredients % to flow away it undefined, to replicate the certainty that if x is on the factor of a million and y is on the factor of ?, then x^y does not would desire to be on the factor of a million.
2016-10-03 07:12:12
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answer #2
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answered by eilermann 4
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If infinity+1 is > and = to infinity, wouldn't that also mean: infinity-1 is < and= to infinity?
I think there is a conspiracy between infinity and zero!
How about 1 and 0. What if 1 contains infinity?
Is zero nul? 0/1=0
Is infinity undifined? 1/0=infinity
2007-06-29 04:17:32
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answer #3
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answered by Yahoo! 5
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use limits
say for the function f(x) = x+1, as x approaches infinity, f(x) approaches infinity, thus infinity +1 = infinity
2007-06-29 03:28:11
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answer #4
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answered by PoseidenNeptuneReturns 4
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the number is so large that the one becomes insignificant thus leading you to infinity+1=infinity
2007-06-29 03:24:37
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answer #5
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answered by LT 4
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a + 0 = a
a + 1*0 = a
( a + 1*0 ) / 0 = a/0
a/0 + 1 = a/0
We have a/0=infinity
infinity + 1 = infinity
2007-06-29 04:03:28
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answer #6
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answered by ? 5
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Infinity is a non-real number and isn't a defined number.
1 is. It's like adding 1 + X = B
Not enough known.
2007-06-29 03:24:35
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answer #7
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answered by Anonymous
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You can't add 1 to infinity. Infinity is a concept that humans are not capabable of really understanding anyway. The way I see it, the only thing that could possibly be infinite is the universe, and who says it is? If you go with big bang theory, than it is not. The concept itself is a trip. you cant add or subtract, multiply or divide. I know you dont want to hear it, but it "just is". who knows if it is even real. like god, or bigfoot, or how many licks it takes to get to the tootsie roll center of a tootsie pop.
2007-06-29 03:31:28
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answer #8
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answered by karateface 2
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Infinity is so large, you don't actually count it as a real number.
2007-06-29 04:14:59
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answer #9
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answered by Elder Price 2
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Infinity is only an imaginary number, not a real number.
2007-06-29 03:32:44
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answer #10
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answered by Anonymous
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