# 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

Staff Emeritus
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).