Polynomial. (1 Viewer)

[Fus Ro Dah]

 

[RealiseNothing]

Is this a proof by contradiction? ie assume there exists such a polynomial and show that the co-efficients or something can not possibly be integers (just an example).

 

[deswa1]

Is this a proof by contradiction? ie assume there exists such a polynomial and show that the co-efficients or something can not possibly be integers (just an example).

Yeah I would assume so- I can't think of another way of doing this. I'll try it tomorrow (inb4 I have no idea what's going on )

 

[asianese]

Just playing around with some numbers

I know it's completely wrong but anyway lol

 

[tomp1612]

Yeah I would assume so- I can't think of another way of doing this. I'll try it tomorrow (inb4 I have no idea what's going on )

if this is a proof by contradiction f***k that I hate that sort of stuff

 

[johnpap]

If such a polynomial existed, then as P has integer coefficients, a-b|P(a)-P(b) -> a-b|b-c, and similarly b-c|P(b)-P(c), so b-c|c-a and then c-a|P(c)-P(a), so a-b|b-c|c-a|a-b..., and then since the modulus of each thing is strictly increasing at each point (Since none of a-b, b-c, c-a can be 0 if a,b,c distinct), we know either a-b=b-c=c-a, or two of them are the same and the third just the negative of whatever value, and in both cases it breaks pretty quickly - i.e. if a-b = -(b-c), a=c which is broken, and if the 3-cycle thing appears then a+c = 2b, a+b = 2c, so 3a/2 + c/2 = 2c, and then a=c so it's broken again cause a,b,c were distinct.

The lemma about a-b|P(a)-P(b) if a,b are integers and P is in Z{x} is pretty useful and also easy to prove, just write out the polynomial in full and note that it splits into many chunks of stuff of the form a^k - b^k, which all are divisible by a-b.

 

[Shadowdude]

if this is a proof by contradiction f***k that I hate that sort of stuff

what

why

 

[asianese]

Omg i got somewhere with the a^k b^k shit?!?! :d