Question 978538
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(a\ +\ b\sqrt{2}\right)\left(a\ -\ b\sqrt{2}\right)]


Is the product of two conjugates.  The product of two conjugates is the difference of two squares.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\sqrt{2}\right)^2\ =\ 2]


by definition of square root, and *[tex \Large 2\ \in\ \mathbb{Z}] by definition of Integer.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ \in\ \mathbb{Z}\ \Right\ a^2\ \in\ \mathbb{Z}],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ \in\ \mathbb{Z}\ \Right\ b^2\ \in\ \mathbb{Z}],


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b^2, 2\ \in\ \mathbb{Z}\ \Right\ 2b^2\ \in\ \mathbb{Z}]


by Closure of Integer Multiplication.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^2,\, 2b^2\ \in\ \mathbb{Z}\ \Right\ a^2\ -\ 2b^2\ \in\ \mathbb{Z}]


by Closure of Integer Subtraction.


Q.E.D.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \