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#### Lexicon Devil

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#### Koboremi

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I think it's supposed to be a number that just uses four 1's, i.e. infinity would not be allowed. I'd say 11^11 but I'm dumb so what do I know.

Edit: I'll change my answer to the factorial of 1111

Lexicon Devil

#### Lexicon Devil

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I think it's supposed to be a number that just uses four 1's, i.e. infinity would not be allowed. I'd say 11^11 but I'm dumb so what do I know.

Edit: I'll change my answer to the factorial of 1111
lol !! You're right. Can't use infinity. So your answer is 1111!

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11!^11!

#### Lexicon Devil

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I think we have a winner. Post your answer on FB if you want.

Edit: Hold the phone.

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#### Lexicon Devil

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Wow !! My brain is hanging upside down.

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∞^∞ :kitteh:

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#### Lexicon Devil

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but you're using infinity, not just four 1s!

Yeah, as I acknowledged to Koboremi I can't use infinity.

11/(1-1) ?? This is an imaginary number.

I want to call you a sweet kitten for some reason. Be awkward to call you this especially in person.

#### Mr. Meepers

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Infinity is not a number that can be used under normal arithmetic operations. Infinity really just means something without "bound". Although, almost paradoxically, some infinities are "larger" than others when we look at set theory. For instance the set of all real numbers between 0 and 1 has a greater cardinality (size) than the set of all integers. Although, every infinite set has a proper subset of the same cardinality (for instance the set of all rationals and its subset, the set of all integers). It may be also interesting to note that the amount of "sizes" infinity can be is also unbounded because a set called the power set always has a larger cardinality than the original set and one can keep taking power sets of a power set.

As for dividing by zero, that is also not allowed because zero is undefined. Why is zero undefined? Because it has no multiplicative inverse.
Now one may be temped to just say infinity should be a number and be zero's multiplicative inverse, but it is not so simple
Let f(a) be zero and let the limit as x->a of g(a) be infinite. Also let f(x) and g(x) be continuous for every point they are defined on, then the limit as x -> a of f(x)*g(x) can be any number, or even infinite
Example: Let n be a non-zero real number and a = 0
1) Let f(x) = |x| and g(x) = n/|x|
Then, for x=/=0 f(x)*g(x) = n i.e. the limit x -> 0 of f(x)*g(x) is n
1) Let f(x) = |x| and g(x) = n/(|x|^2)
Then, for x=/=0 f(x)*g(x) = n/|x| i.e. the limit x -> 0 of f(x)*g(x) is infinity (or negative infinity if n is negative)

We also get contradictory results if we allow division by 0:
i.e.
2=2
2= 2 + a - a
2/(a-a) = [2/(a-a)] + 1
2/(a-a) - [2/(a-a)] = 1
0 = 1

So don't divide by zero

I'm Serious!! :shocked:

#### ae1905

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Yeah, as I acknowledged to Koboremi I can't use infinity.

11/(1-1) ?? This is an imaginary number.

I want to call you a sweet kitten for some reason. Be awkward to call you this especially in person.
imaginary numbers are defined by the square root of -1

#### ae1905

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Infinity is not a number that can be used under normal arithmetic operations. Infinity really just means something without "bound". Although, almost paradoxically, some infinities are "larger" than others when we look at set theory. For instance the set of all real numbers between 0 and 1 has a greater cardinality (size) than the set of all integers. Although, every infinite set has a proper subset of the same cardinality (for instance the set of all rationals and its subset, the set of all integers). It may be also interesting to note that the amount of "sizes" infinity can be is also unbounded because a set called the power set always has a larger cardinality than the original set and one can keep taking power sets of a power set.

As for dividing by zero, that is also not allowed because zero is undefined. Why is zero undefined? Because it has no multiplicative inverse.
Now one may be temped to just say infinity should be a number and be zero's multiplicative inverse, but it is not so simple
Let f(a) be zero and let the limit as x->a of g(a) be infinite. Also let f(x) and g(x) be continuous for every point they are defined on, then the limit as x -> a of f(x)*g(x) can be any number, or even infinite
Example: Let n be a non-zero real number and a = 0
1) Let f(x) = |x| and g(x) = n/|x|
Then, for x=/=0 f(x)*g(x) = n i.e. the limit x -> 0 of f(x)*g(x) is n
1) Let f(x) = |x| and g(x) = n/(|x|^2)
Then, for x=/=0 f(x)*g(x) = n/|x| i.e. the limit x -> 0 of f(x)*g(x) is infinity (or negative infinity if n is negative)

We also get contradictory results if we allow division by 0:
i.e.
2=2
2= 2 + a - a
2/(a-a) = [2/(a-a)] + 1
2/(a-a) - [2/(a-a)] = 1
0 = 1

So don't divide by zero

I'm Serious!! :shocked:
infinity doesn't obey the usual rules finite numbers do, but it is the unbounded limit of the real number line--ie, can be put alongside real numbers--and can be expressed as the limit of 1/x as x-->0, which for infinitesimally (vanishingly) small x is essentially infinitely big--ie, infinity

#### Lexicon Devil

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#### EvilDrFrog

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lim[SUB]x->1[/SUB] 11/(1-x)

Equals infinity. Problem solved.

#### Lexicon Devil

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lim[SUB]x->1[/SUB] 11/(1-x)

Equals infinity. Problem solved.
Frog math. :-/

#### Mr. Meepers

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infinity doesn't obey the usual rules finite numbers do, but it is the unbounded limit of the real number line--ie, can be put alongside real numbers--and can be expressed as the limit of 1/x as x-->0, which for infinitesimally (vanishingly) small x is essentially infinitely big--ie, infinity
True, but my comment was directed more at things like infinity ^1111 or 1111^infinity lol
Although, I would use limit of 1/|x| as x ->0 because it is + infinity coming from either side on the limit

And the riddle does say "number", so an answer of infinity would be debatable if it were acceptable. Clearly the riddle needs more clear rules This riddle is not "well defined"

The font only goes to 7 here maybe just "1" at font 111

Oh, I got it:

+

And I still have three 1s to spare

Edit:
1
1
1
1

They're stackable ... I believe I have the "highest" numbers

Edit: Edit:
f: R --> {1}
x --> 1

sup (f[SUP]-1[/SUP](1))

(Basically sup ({x|f(x)=1}))

f: (0,1) --> R
x --> 1/x

sup ({f(x)| x is an element of (0,1)})
And I have a "1" to spare

sup (R) where R is the set of real numbers
There: 4 "1"s to spare

#### Lexicon Devil

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@Mr. Meepers
You're too funny. Stackable :laughing: Font 111 :laughing:

Well, it's quite obvious that my understanding of math is low. ;-)

All I know is that Infinity is used in set theory in some calculus problems. I think.

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