Homework Help: Please check my (simple) proof. Skeptical of its simplicity

1. Sep 5, 2011

Abraham

x,y are in R. Suppose x2+y2=0. Prove that x=0 and y=0.

My proof:

Suppose x$\neq$0, y$\neq$0. Then by the field axioms, both x2 and y2 are strictly positive, and so is their sum. This is a contradiction, since we supposed that their sum = 0.

Thus, x=0, and y=0.

This problem and proof seem so simple, I think there may be something wrong with it.

2. Sep 5, 2011

stallionx

What specific geometric shape does it define in the Cartesian ?

3. Sep 5, 2011

dynamicsolo

Yes, you're not quite there: you have shown that x and y can't both be zero. Could either one be zero? (Well, of course, they can't, but you still need to include something about this in your argument.)

4. Sep 5, 2011

Abraham

Is it a circle? The hw problem is for an intro to analysis class though, so I wonder if a geometric argument / proof would be accepted.

To dynamicsolo: Thanks, that's what I was looking for. I felt like I was missing something.

5. Sep 6, 2011

Syrus

The structure of your method of proof is incorrect in general. You are attempting a proof via contradiction. The negation of a conjunction is not simply the conjunction of the negations of its conjuncts. Finding and using the correct form of your new premise (assumed as a result of contradiction) will lead you to a more traditional and sound argument.

6. Sep 6, 2011

Abraham

Hi Syrus. Do you mind clarifying? I don't think I understand what a traditional argument is. What makes a sound proof by contradiction? So far, I show, by contradiction that:

1.) x $\neq$ 0 ---> x2+y2 $\neq$ 0
2.) y $\neq$ 0 ---> x2+y2 $\neq$ 0
3.) x, y $\neq$ 0 ---> x2+y2 $\neq$ 0

Thus, x=0, and y=0. How do I improve the argument? Thanks.

7. Sep 6, 2011

HallsofIvy

Actually, that's not simple enough! You don't need line (3).
Although I would add more words:
"1) if $x\ne 0$ then $x^2> 0$. $y^2\ge 0[/tex] so [itex]x^2 +y^2> 0$. Contradiction"

8. Sep 8, 2011

Syrus

Well i meant that the negation of (x = 0 AND y = 0( is (x =/= 0 OR y =/= 0). At least that's the way I would look at it and proceed (via proof by cases).