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I need to explain this to someone.

I know obviously the expanded form gives you $x^2 + 2xy + y^2$ but technically don't the individual exponents multiply to give $x^2 + y^2$

I might say I'm looking for an interesting geometrical explanation for this.

Any help is appreciated!

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    http://www.math.brown.edu/~banchoff/Beyond3d/Images/chapter2/image18.jpg – 2017-01-14
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    No. However there are settings where $(x + y)^2 = x^2 + y^2$ does hold (though the meaning of $+$ and $\cdot$ is not as you know it from the ring of real numbers). See e.g. [tropical geometry](https://en.wikipedia.org/wiki/Tropical_geometry) – 2017-01-14
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    Oy, the answer is oi. – 2017-01-14
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    Please, explain that you're either joking, or that you're trying to convice a stubborn student of yours about that identity and you want a geometric argument... Or, that you're the sibling of TripleA, who is a maths graduate... – 2017-01-14
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    @Pythagoricus The second of those options haha :) – 2017-01-14
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    Why? Simply because it isn't. Pick any two numbers, call them X and y, and compute $(x+y)^2$ and $x^2+y^2$. Are the two results equal? – 2017-01-14
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    I badly worded my question, I was just looking for a geometrical explanation to why, to show someone else. – 2017-01-14
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    @TripleA Thank God! I'll survive this scary night! My opinion: it's a unique oportunity to teach your student about the power of Logic. I mean, ask them to give you a rigorous justification of why they think the identity is false, and show them that they cannot back what their claim with reason. It's one of the best lesson you can teach a student! – 2017-01-14
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    In my opinion it is important to stress the requirement of the commutative property to hold. $(A+B)^2=A^2+AB+BA+B^2$ is the more general case. – 2017-01-14
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    @Pythagoricus I did, but then he asked me the same, hence my question – 2017-01-14
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    @TripleA Good luck! But don't forget, it's much better to have somebody who objects when they don't understand something than one who just parotes what the teacher says... – 2017-01-14

4 Answers 4

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Source: https://letsplaymath.files.wordpress.com/2008/07/picture24.png

enter image description here

Does it explain your doubt?

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If $x=y=1$, then it should be obvious that

$$(1+1)^2=2^2=4$$

$$1^2+1^2=1+1=2$$

So it should be trivial that

$$(1+1)^2\ne1^2+1^2$$

And it cannot hold in general if it fails for at least one case.

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Actually, it is true - sometimes. For fields of characteristic $2$ we have $(x+y)^2=x^2+y^2$. In general this has the nice name freshman'as dream, and this name indicates that we cannot expect it to be true in general (it would be a dream).

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    But the dream ends too soon. `;-)` – 2017-01-14
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I'm going to assume that you're familiar with the expansion of$$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd\tag1$$ With $(x+y)^2$, since the definition of an exponent is that $a^n=\underbrace{a\cdot a\cdot a\cdots a}_{n\text{ times}}$. Hence,$$(x+y)^2=(x+y)(x+y)=x(x+y)+y(x+y)=x^2+xy+yx+y^2\\=x^2+2xy+y^2$$