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Consider the following implication, $x,y \in \mathbb R \wedge x \lt 0 \implies \nexists y$ such that $x=y^2$. Question asks to use contrapositive, so here is my proof:

Let $x=y^2$ (since it's negation of conclusion). I want to show that $x \ge 0$.

So from new hypothesis we know that x is positive real number greater than or equal to 0, since $x=y^2$ (x is equal to y square) this means the new conclusion is correct. So contrapositive is true meaning implication is true.

Is my proof correct, or am I missing something?

Thanks!

Edits: Could someone please give me correct proof for this problem? Thanks.

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You seem to simply be asserting what you want to prove, instead of actualy proving it, when you say "we know [the conclusion] is true, since [our hypothesis holds.]" I would not consider this a good proof. Instead, what I would look for is something along the lines of a proof: "suppose $x=y^2$; if $y\lt 0$, then...; if $y=0$, then...; and if $y\gt 0$, then...; in conclusion if $x=y^2$, then $x\geq 0$." –  Arturo Magidin Aug 3 '11 at 5:58
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Are you interested in having the correct structure for the proof, or the correct proof? I the second, then I have not seen where you showed that it follows that $x \leq 0$ , but you may be able to, by considering what happens when y is positive and then when y is negative. –  gary Aug 3 '11 at 5:59
    
Your proof is circular. You want to show $P \Longrightarrow Q$. You then assume $\neg Q \Longrightarrow \neg P$ and conclude $P \Longrightarrow Q$. –  JavaMan Aug 3 '11 at 5:59
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As Arturo and DJC pointed out, you are just asserting what you want to prove. Think about it: How can you possibly prove anything about the real numbers without using any facts about the real numbers? –  Braindead Aug 3 '11 at 6:16
    
You mean something like assume $x=y^2$ means for $y>0, x>0$ and for $y<0$, $x>0$ and for $y=0 x=0$ which means for $x=y^2 $ implies $ x>=0$? (got hint from the first comment) –  Mayumi Aug 3 '11 at 14:10

4 Answers 4

up vote 2 down vote accepted

:-) look what youve done: you just put "new hypothesis" and said "if the new hypothesis is true, then the new hypothesis is true". And from this you concluded that the implication is true :-)

The correct proof would depend on the axiom set. Suppose we know that if $a \ge b$ and $c > 0$, then $ac \ge ab$, that $a \cdot 1 = a$, $a + 0 = a$, $a + (-a) = 0$, $a + (-b) = a - b$, and $a \cdot (b + c) = ab + ac$.

You want to prove that for $y \in \mathbb{R}$, $y^2 \ge 0$

1) if $y \ge 0$, then $y \cdot y \ge 0 \cdot y$, thus $y^2 \ge 0$, QED

2) if $y < 0$, then it is more complicated, first prove that $-y = (-1) y$:

$0 = (-1 + 1) y$, thus $0 = (-1) y + y$, thus $-y = (-1) y$

Then prove that $(-1)(-1) = 1$: we know that $1 + (-1) = 0$, thus $-1 \cdot 1 + (-1)(-1) = 0$, and thus $(-1)(-1) = 1$

Finally, from 1) we know that $(-y)^2 \ge 0$, so:

$0 \le (-y)^2 = ((-1) y)^2 = (-1)(-1) y^2 = 1 \cdot y^2 = y^2$

QED

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I think you've got the wrong end of the stick, Tomas. If you look at the question, then you will see that Mayumi is not trying to prove that for all $y \in \mathbb{R}$, $y^2 \ge 0$. This fact is important to the proof, yes, but can probably be assumed. The question is in fact asking to prove that if $y$ is negative, that there is no $x \in \mathbb{R}$ such that $x^2=y$. This requires the contrapositive to prove. Mayumi's solution is completely correct - if there is some $x$ such that $x^2=y$, then $y$ is positive; i.e., negation of conclusion implies negation of premise. QED. –  Donkey_2009 Mar 25 '12 at 11:00

forall x, x^2 > 0

therefore (by taking the contrapositive)

there does not exist x, x^2 < 0

now you can introduce 'y' if you want but it's redundant.

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Proof by contradiction.

Suppose not. Therefore, $x = y^2$ where $x,y \in \mathbb{R}$ and $x < 0$. Since $y \in \mathbb{R}$, $y^2 \ge 0$ and thus $ x \ge 0$ (by assumption). But this contradicts the fact that $x<0$ , which implies that $x \neq y^2$.

Hence, since $x \neq y^2$ for $x,y \in \mathbb{R}$ and $x < 0$, $\nexists y$ such that $x = y^2$.

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I don't know what everyone's going on about, as your proof is completely correct. You could have been a little clearer in your wording, though.

You want to prove that $x,y \in \mathbb R \wedge x \lt 0 \implies \nexists y$ such that $x=y^2$, using the contrapositive.

Quite rightly, you start by assuming that $\exists y$ such that $x=y^2$; i.e., the negation of the conclusion you wnat to prove. You then set out to prove the negation of the premise, namely that $x \ge 0$. This is exactly how the contrapositive works.

You prove this by noting that $x$ is a square and must therefore be positive. So you win!

Don't listen to these people telling you that your proof is wrong. I'm not quite sure why so many people think that it's wrong, but they're the ones who are wrong, not you.

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I guess if he already knows that all real squares are non-negative then he can assert this. But since this is almost exactly the same thing as what he is trying to prove, I think it's pretty safe to assume that he cannot use that fact. Otherwise, yes, it is a trivial proof because it's just a simple logical manipulation. –  asmeurer Jun 17 '12 at 3:14
    
My guess was that the point of the question was to teach proper use of the contrapositive, rather than to provide a formal proof that squares are non-negative. Perhaps I'm wrong, though. –  Donkey_2009 Jun 20 '12 at 12:11

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