I'm going through the book Calculus by Michael Spivak, not sure on how to go about the first problem. If anyone can just go through it with a solution I think I can handle myself from there. Thanks.
Prove the following:
$x^2−y^2=(x−y)(x+y)$
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1Try using the FOIL rule. – ncmathsadist Dec 29 '16 at 02:29
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1$(x-y)(x+y)=(x-y)x+(x-y)y=\dots$ – JMoravitz Dec 29 '16 at 02:29
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Oh yes distribution! Thank you @JMoravitz – Alexander John Dec 29 '16 at 02:30
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If I recall correctly the first chapter of Spivak's Calculus is devoted to the axioms of the real numbers and they are listed as P1, P2, ... Just use them successively to expand the RHS – Dec 29 '16 at 02:31
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1I wasn't sure if I was doing it right thank you @mathbeing – Alexander John Dec 29 '16 at 02:33
4 Answers
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$(x-y)(x+y)=(x-y)(x)+(x-y)y= (x^2-xy)+(xy-y^2)=x^2-y^2+ ( xy-yx)=x^2-y^2+0=x^2-y^2$. (note that near the end we used $xy=yx$).
Asinomás
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Does it matter with side of the equation you choose to start proving from? @jorge – Alexander John Dec 29 '16 at 02:35
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no it doesn't, because you proved that the two things are equal either way. – Asinomás Dec 29 '16 at 02:38
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So, you'd use the distributive law and cancellation by additive inverse: $(x-y)(x+y) = x(x+y) - y(x+y) = x^2 + xy - xy - y^2$
Amin Idelhaj
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2yeah, and commutativity of both addition and multiplication is used various times. – Asinomás Dec 29 '16 at 02:32
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Treat $x^2-y^2=0 $as a quadratic equation in variable $x$. Use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ so that
$x=\frac{0\pm\sqrt{0+4y^2}}{2}=\pm y$
So $x^2-y^2=(x+y)(x-y)$
Nitin Uniyal
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Alternately, $(x^2-y^2) = (x-y)^2-2y^2+2xy = (x-y)^2 -2y(y-x)=(x-y)^2 +2y(x-y) =(x-y)[(x-y+2y)]=(x-y)(x+y)$