Part 3 assumed that f(1) = 1, and ignored the possibility of f(1) = 0. A mistake! A mistake! Terrible! Terrible! Ignore it in the case that f(1) = 0 ! Ignore it! Ignore it!
Part 6: for all rationals r, we either have that ∀r f(r) = 0 or we have that ∀r f(r) = r.
Suppose we are in the case that f(1) = 0. Then it was shown that for all integers n, f(n) = 0. So for a rational, r = p/q, we have that:
>>217
I'm trying to show that f is the identify function in one case, but my only assumptions are given in >>192 as P1 and P2.
Sorry >>190-san. While your bot was most disruptive, it was also creative in its own right.
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Anonymous2013-02-11 7:36
Try proving it isn't the identity function, i.e. assume f(n) ≠ n, then continue the proof by induction as you were.
No idea how it would go, but it's worth a try?
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Anonymous2013-02-11 7:50
I'm just wondering: have you guys ever craved cock so badly that you found yourself running around outside, howling at the moon for it? Literally ROARING at the top of your lungs, wanting nothing less than a dick's head churning against your glottal stop?
Tell me I'm not alone.
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Anonymous2013-02-11 22:48
>>219
dont remember euclid needing your identity function...
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Anonymous2013-02-12 0:38
>>221
don't remember patrick bateman checking my TRIPS
1. If f(1) = 0, then f is the zero function.
2. If f(1) is not 0, then f is the identity function on the rationals.
If f has to be continuous, then (2) implies that f is the identity on the reals as well, but without continuity I can't connect the rationals to non-rationals so easily.
Part 7: if f(1) is not zero, then f is id on all real nth roots of all rationals.
Let n be a power and r a non-negative rational.
Let b be the unique real nth root of r, b = root(r,n)
Since f(b) multiplied by itself n times is equal to r, we have that f(b) is an nth root of r. Because real roots are unique, this implies that f(b) = b. XD
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Anonymous2013-02-18 12:34
>>225 this implies that f(b) = b. XD
That's not even funny, fagshit.
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Anonymous2013-02-18 13:18
dont remember euclid needing your identity function.
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Anonymous2013-02-18 15:55
wut?
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Anonymous2013-02-18 16:04
NEETkita Le Hikikomori
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Anonymous2013-02-19 2:00
Well, it can be shown by induction of each of the parts that any number that can be formed by a finite number of additions, subtractions, multiplications, divisions, and square roots, applying the said operations in any order, then f fixes this number. So that covers quite a few numbers, but I really doubt that every real number can be expressed in this way. I could prove that by showing the set of all finite length algebraic expressions using sums, differences, multiplications, quotients, and roots, to be countable. I guess that is obvious, since the set of such algebraic expressions would correspond to a subset of the language of all strings containing '( ) + * - / ^ 1' which is a countable set.
So no, I don't have the reals yet. I think continuity is needed. I'll try to produce a counter example.
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Anonymous2013-02-19 2:09
>>230
Prove that "induction" is a provable way prove anything.
>>231
Give me an expression using n applications of the said operations, and I'll write you a proof that uses n lines, using the parts already shown at each step.