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For $x_1, x_2 \in \mathbb{R}$ there is the rule: $(x_1\times x_2)^2=x_1^2\times x_2^2$. How can I prove, that this rule doesn't apply for: $(x_1+x_2)^2$?

Arthur
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3 Answers3

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$$(x_1+x_2)^2=x_1^2+x_2^2$$ $$x_1^2+x^2_2+2x_1x_2=x_1^2+x_2^2$$ $$2x_1x_2=0$$ $$x_1=0 \text{ or }x_2=0$$

Teoc
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this statement doesn't need proof. It just needs a counterexample:

put $x_1=-x_2=1$

souran
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  • Your answer would have been correct if the logical form of the statement was $$(\exists x_1\in\mathbb{R})(\exists x_2\in\mathbb{R})((x_1+x_2)^2\neq x_1^2+x_2^2).$$ But here we have $\forall x_1,x_2$. – Workaholic Feb 08 '15 at 19:27
  • #workaholic the second statement that you have stated is also False. – souran Feb 08 '15 at 19:29
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    mmahdit your counterexample does not work because it satisfies $$(1-1)^2\neq 1^2+(-1)^2$$ – Teoc Feb 08 '15 at 19:30
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    it's a counterexample for this : $(x_1+x_2)^2={x_1}^2+{x_2)^2$ – souran Feb 08 '15 at 19:56
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    But the question asked to prove $(x_1+x_2)^2\neq x_1^2+x_2^2$ – Teoc Feb 08 '15 at 20:02
  • you have mistaken the statement with an equation and you have treated it like finding two numbers that satisfy the equation. and your statement is wrong and the proof of its wrongness is a counterexample. When you want to prove that a statement is not true, logic says you should only provide an example for which the statement does not hold. This example as you know is called "Counterexample". When someone says all black people have mustache you don't need to collect all the population of black people without mustache, You just need to show ONE black man without mustache. just ONE...ok? – souran Feb 09 '15 at 09:06
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As the integers are contained in the real numbers the statement implies that the sum of any two square numbers is also a square number. $4+9=13$ and $13$ is not square leading to a contradiction. It's fun to see the consequences if it were true!

Karl
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