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Inducting on P means, you need to show that the statement is true for all \(P\in \mathbb{N}\).

hm

why not simply take base case as \(P=1\)

that doesnt have diameter 3

what diameter

oops ignore that im mixing up questions now

okay p=1

so any tree with P>1 is 2-chromatic

|dw:1442977242648:dw|

i feel like this induction is not complete

thats not induction, just showing it for convincing ourselves first

okay i see what u mean

|dw:1442978241263:dw|

oh i see if we remove A then we cant use that argument

yeah there should be a better way to approach this... this removing and adding thingy may not work

i can do 2 cases

i think it might get messy..

ya theres gotta be a better way

oh wait i can!

nvm i cant

|dw:1442978612649:dw|

okay i have an idea

say A is connected to multiple like u did there

u must be able to right!!

this is my idea in a diagram

|dw:1442979234709:dw|

|dw:1442979350004:dw|

and that case is very easy to prove

would this work?

that looks rather complicated but the construction definitely works!

ya that looks like a lot to write in this one proof lol

i have a better idea

okay go on

it is simple

lets just make a choice of which vertex we remove

oh omg lol xD

ya okay that is done

simply remove a "leaf vertex", the problem is solved :)

yep

how do you know a leaf vertex always exists ?

looks neat !

thanks a lot for your help