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My question isn't specifically about the problem below, but about coloring proofs in general. From what I've seen about coloring proof problems thus far, I've noticed that we form a conditional and find a contradiction, i.e we can't cover a checkerboard with the 2 diagonal corner pieces removed with $2$ x $1$ pieces, or find cases where we could fit certain pieces onto a board. But I'm unsure how these conditionals are formed since the proofs I've seen usually don't do so explicitly. For the problem below, would we write if we could fit $250$ $4$x$1$x$1$ bricks in a $10$x$10$x$10$ box with the coloring as such, then each brick would contain each color. Or do we write if we could fit these pieces into the box then each brick would contain 4 different colors?

Thanks

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john fowles
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Your two offered conditionals don't really mean anything, as they both contain "then each brick would contain each color", but the if part of the statement does not provide the reasoning why. So here is here is the general idea to coloring proofs by what I more think of as contrapositive:

First we construct a coloring such that there is some result such as "each brick contains each color".

Then with this coloring, and these bricks, then we derive some result about the colors in any tile-able shape. In your case it is that anything tile-able must contain equal numbers of each color. To be clear, this looks like $$\text{Tile-able} \implies \text{Condition}$$

Finally, what we really do here is take the contrapositive. If the shape that we are attempting to tile does not satisfy the condition that tile-able shapes must have, then it must not be tile-able by that brick, that is, $$\text{Not Condition} \implies \text{Not Tile-able}$$ which is the contrapositive of our first statement which we derived, and is thus true.

I hope that clears some things up, and if anyone has quibbles about this contrapositive vs. contradiction thing, please refer to this excellent post Proof by contradiction vs Prove the contrapositive.

Isaac Browne
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