Knot Theory, 3colourbility of knots
Decide the 3colourability of each of the knots 7_5, 7_6, 7_7: In each case, either give a valid
3colouring or prove that no 3colouring exists.
Use the Rolfsen Knot Table
Answer
 The questioner was satisfied with and accepted the answer, or
 The answer was evaluated as being 100% correct by the judge.
1 Attachment

Thanks I understand 7_7 however what I did for 7_5 and 7_6 is that I proved it by trying to draw the strands with only 3 colours however I said that since the under/over crossing strands must be different it must use more than 3 colours, do u think that is still a valid proof ?

If you formalize it correctly, it is, but I think that formalizing it correctly is essentially equivalent to doing what I did here. You can't just say "I tried and it didn't work", you have to do a casebycase analysis, explain why forced choices are actually forced, and then make sure that your cases are exhaustive. It is doable, but you need to be careful. I find this linear algebra approach more streamlined.

Also under/over crossings don't have to be different, they can be the same color if the third strand in the crossing also has the same color, that's the tricky part I think.

on wikipedia it says 7_5 is tricolorable?


Ok makse sense I will look over what I did!

on wikipedia it says 7_5 is tricolorable?


oops i misread it sorry
 answered
 751 views
 $40.00
Related Questions
 Generalization of the Banach fixed point theorem
 Undergrad algebraic topology proof
 Let $(X, \cdot)$ be a normed space. Let $\{x_n\}$ and $\{y_n\}$ be two Cauchy sequences in X. Show that the seqience Show that the sequence $λ_n = x_n − y_n $ converges.
 Prove that $S \subseteq X$ is nowhere dense iff $X\overline{S}$ is dense.
 Pathwise connected
 Show that ${(x,\sin(1/x)) : x∈(0,1]} ∪ {(0,y) : y ∈ [1,1]}$ is closed in $\mathbb{R^2}$ using sequences
 Given locally limited $f:[0,1]→\mathbb{R}$, show that $Graph(f)$ is closed in $\mathbb{R^2}$ ⟺ $f$ is continuous using sequences
 Prove that $p_B :\prod_{\alpha \in A} X_\alpha \to \prod_{\alpha \in B} X_\alpha$ is a continuous map