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Let's say that $X$ is a parallelogram with vertices that have integer coordinates, how could I prove that $X$'s area is an integer?

The vertices are $0, A, B$ and $A + B$. How would I do this?

  • In the future, please do not remove relevant context/working from your post. I don't know why you did that, and it is likely the reason why this post has accumulated three close votes (and nearly a fourth by me). MSE generally discourages contextless posts (where context can include your motivation, understanding, and attempts) - your edits to the question rendered this precisely such a post. (That's not to say your original post was stellar but you definitely made it worse in the eyes of MSE.) [cont.] – PrincessEev Mar 22 '19 at 04:06
  • I have done some edits to your post to account for this - I included the original post it was written through a rollback (and after an accidental derp on my behalf restored the original title). The only edit that was retained was Andreas Blass' tag edits. The full revision history, for anyone who's curious -- https://math.stackexchange.com/posts/3146771/revisions – PrincessEev Mar 22 '19 at 04:07

2 Answers2

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The area of the parallelogram with two sides defined by the position vectors $A=(a,b)$ and $B=(c,d)$ from the origin is given by the magnitude of the cross product of the vectors: $$|A×B|=|(0,0,ad-bc)|=|ad-bc|$$ Since all of $a,b,c,d$ are integers here, the area $|ad-bc|$ is also an integer.

Parcly Taxel
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If $A=x_1 +iy_1$ and $B=x_2+iy_2$ then

$A\bar B = (x_1+iy_1)(x_2-iy_2) = (x_1x_2 +y_1y_2) + i(x_2y_1-x_1y_2)$

so $|Im(A\bar B)| = |x_2y_1-x_1y_2|$, which is the area of the parallelogram.

gandalf61
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