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Say if $x_0$ and $y_0$ are integers. I have $(x_0+y_0)(x_0+y_0)$. I know this is equal to $x_0x_0 + 2x_0y_0 + y_0y_0$, but I want to prove this using the mathematical axioms.

What would be the best thing to do. I was considering using the distributivity law for integers to get: $x_0(x_0+y_0)$ + $y_0(x_0+y_0)$. Then I would have $x_0x_0+x_0y_0 + y_0x_0 + y_0y_0$, then using commutativity of multiplication $x_0x_0+x_0y_0 + x_0y_0 + y_0y_0$. I could then just assume $x_0y_0 + x_0y_0 = 2x_0y_0$ ?, giving me the correct final answer ?

I am unsure on how to start this as I am sure I am missing a step or two. I am also unsure whether I can assume $x_0y_0 + x_0y_0 = 2x_0y_0$. Thanks : )

Bill Dubuque
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    By distributivity $$a+a= 1 \cdot a + 1 \cdot a = (1+1) \cdot a = 2a$$ – Crostul Oct 28 '22 at 10:42
  • Okay thanks. I guess I should be more rigorous than taking the risk of being less rigorous. Any idea for my first question about how to start it ? @Crostul –  Oct 28 '22 at 10:46
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    You can save a lot of indices by writing $x=x_0$ and $y=y_0$. – Dietrich Burde Oct 28 '22 at 11:37
  • @DietrichBurde For this question I have to do it like that though. Can you let me know how to start off then ? –  Oct 28 '22 at 11:54
  • No, you don't have to do it like that. Other people have answered it writing $a,b,c,d$. Rewrite it with $a=c=x_0$ and $b=d=y_0$. – Dietrich Burde Oct 28 '22 at 12:39

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