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I have the problem:

Prove that $(x + y)^2 \not= x^2 + y^2$

I have answered it with something like this:

$$(x+y)^2 = x^2 + 2xy + y^2$$

$$(x^2 + 2xy + y^2) - (x^2 + y^2) = 2xy$$

$$\therefore (x + y)^2 = x^2 + y^2 \Rightarrow 2xy = 0$$

$$\therefore (x+y)^2 = x^2 + y^2 \Rightarrow x=0 \lor y=0$$

(Apologies if I'm using symbols incorrectly here, I'm very new to this sort of thing, and as I'm teaching myself I have no one to point these sorts of things out to me)

My problem is that, as far as I can tell, this proves that $(x + y)^2 = x^2 + y^2$ for at least some combination of values for $x$ and $y$. So I suppose my question is, for this kind of proof question, am I being asked to prove that the two statements are not equivalent, or that they are never equal?

Here's a picture of the problem: enter image description here

Blue
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OliverRadini
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  • Just found something of a duplicate here https://math.stackexchange.com/questions/3240691/proof-by-contradiction-to-show-two-equations-are-not-equal which seems to confirm what I'm saying? I'll leave this up for now as the questions are slightly different to my mind, but I'm probably biased – OliverRadini Jan 14 '21 at 20:19
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    Can you put a picture of the original context/problem? It is probably asking to prove that this is not the case for every choice of reals x and y (since there are cases where it is true, like x = y = 0, per what univalence said). In which case you can just take x = y = 1 and be done with it. – Countable Jan 14 '21 at 20:20
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    This arguably seems like a very weird wording, and usually, to be asked to prove $x \not\equal y$ means proving $\forall x.\forall y. x \not\equal y$, which is, for some reason, not the case here – univalence Jan 14 '21 at 20:21
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    The meta-language here is not fully understandable - they ask you to prove a certain expression, which means most of the times that it is for all x, y. But then, as you showed, for $x=0$ and $y=$basically anything, this actually is a false argument (saying that $y^2 \neq y^2$) And thus the command "to prove" fails here. – CSch of x Jan 14 '21 at 20:31
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    You are hundred percent right in your solving. The correct wording should go along this line "prove that A=B is false in general" or "is always false" or "is false except for a finite number of values", etc... As currently stated, the problem is wrong unless somewhere some properties for $x,y$ prevent them for being zero (e.g. strictyl positive for instance). – zwim Jan 14 '21 at 20:32
  • Thanks for all the comments, really helpful to have input one this from people who know what they're talking about. @countable I've added an image of the question now. – OliverRadini Jan 14 '21 at 20:33
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    @univalence: MathJax hint: we have \neq for $\neq$ instead of \not \equal because it comes up so often. You need to leave a space after \equal – Ross Millikan Jan 14 '21 at 20:34
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    @RossMillikan Thanks, I wasn't aware! – univalence Jan 14 '21 at 20:35

3 Answers3

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The statement $(x+y)^2\ne x^2+y^2$ contains "free variables" and whether it is true or not depends on the values of $x$ and $y$.

If $x=y=0$, then this statement is clearly false.

It is correct to say that "there exist real numbers $x$ and $y$ such that $(x+y)^2\ne x^2+y^2$".

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    Yes, exactly: it would have been so much better if the question had been something like, “Show that the statement ‘for every $x$ and $y$, $(x+y)^2\ne x^2+y^2$’ is false.” The question, as given, is defective because of the total lack of quantification. – Lubin Jan 14 '21 at 21:05
  • Thank you for the answer, certainly clarifies things for me! – OliverRadini Jan 15 '21 at 12:11
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You are correct. The statement ought to be $$\text{ Prove that } (x+y)^2\not\equiv x^2+y^2$$

For which it suffices to simply give a counterexample: $4=(1+1)^2\neq1^2+1^2=2$.

The technical distinction between $=$ and $\equiv$ is often overlooked, especially in high school courses, but it is very important. $=$ means it works for a specific value or set of values, $\equiv$ means it works for all values.

Rhys Hughes
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Without further information there is a couple ways to interpret the problem statement.

  1. Perhaps it is to prove that $(x+y)^2 \neq x^2 + y^2$ fails to be an identity. Which is to say there exists values for which it fails. Thus we prove it is not an identity by giving a counter-example to $(x+y)^2\neq x^2 + y^2$.

  2. We could interpret it to be asking us to prove $(x+y)^2 > x^2 + y^2$ or $(x+y)^2 < x^2 + y^2$. However, this also is case dependent. So again would fail to be an identity.

My guess would be that it is asking for the first interpretation I have listed above.

oliverjones
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