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# How do you prove that if x^2 + y^2 = 0, then x = 0 and y = 0?

The proof is so obvious, but I can't seem to figure out the proper steps to know how to do it.

If anybody has an elegant rigorous proof as to why that is true, please post it. No arbitrary verbiage, please. Actual proof with deductive reasoning.

This is how far I've gotten in the proof, but still don't know where to go.

x^2 >= 0

y^2 >= 0

Therefore

x^2 + y^2 >= 0

Not sure how to make the leap of logic that x = 0 and y = 0. Sure, I can say it in words and know how true it is in words, but I'd rather see the algebraic steps.

I'd also prefer not to resort to using imaginary numbers, since I used this fact in a proof involving them.

### 9 Answers

- Professor MaddieLv 41 decade agoFavorite Answer
Suppose x, y are real numbers.

Proof by contradiction.

Suppose that x^2+y^2=0, but x ne 0 or y ne 0 (ne means not equal).

Without loss of generality, let x ne 0. Then x^2 is an element of R+ since this is the range of the function for all nonzero values of x.

This implies that x^2 = - y^2 > 0, hence y^2 < 0. Contradiction since y^2 is also in R+. Hence contrary to my assumption, x=0 and y=0.

If you need to justify that the range of these functions is all nonnegative real numbers, you can use a calculus argument with derivatives, but I don't think this is necessary because they are such elementary functions.

- 1 decade ago
Any number when squared gets a positive value.

Now, if you square a non-zero number 'x', and then square an another non-zero number 'y', and add both of them, you are sure to get a positive value that is greater than zero. This is because you have squared both x and y, and so they are now positive.

But in this case, the value on the right hand side is zero. This can only happen when both x and y are zero. In place of zero, if you have other numbers for x and y, you are sure to get positive numbers, whose sum will again be a positive number that is greater than zero. Therefore, for the condition x^2 + y^2 = 0, x and y both should be equal to zero.

If you want a step wise proof, try this:

Take x=0, y=0. Substitute in the equation. We get x^2 + y^2 = 0.

Now take x=1, y=0. Substitute in the equation. We get x^2 + y^2 = 1.

Now take x=1, y=1. Substitute in the equation. We get x^2 + y^2 = 2.

And go on taking all positive and negative values for x and y. For no value of x and y except zero, you will get the value as zero.

- Pi R SquaredLv 71 decade ago
Hi,

If x and y are real numbers, but are not equal to zero, then x^2 and y^2 will both be positive numbers, and the sum of 2 positive numbers will never equal zero. If one was zero but the other was some other real number, then x^2 + y^2 would still be positive. So the only way using real numbers to get an answer of zero would be if both x and y equaled zero, since 0^2 + 0^2 = 0.

Remember that for complex numbers if x = i^2 and y = i, that x^2 + y^2 = (i^2)^2 + (i)^2 = i^4 + i^2 = 1 + -1 = 0

- Mein Hoon NaLv 71 decade ago
we are dealing with real numbers otherwise this is not true

x =1 y = i satisfies it

now x = 0 and y = 0 satisfies the condition

we shall prove by contradiction

let x not be zero

thern x^2 >0

so x^2+y^2 > y^2 > 0 which is not true

so x not = 0 is not possible

similarly y not = 0 is not possible

QED

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- 1 decade ago
x^2+y^2 = 0, so x^2 = 0 & y^2 = 0 ( as x^2 and y^2 cannot b less than 0) , so x=0 & y = 0

- James ChanLv 41 decade ago
you have got the right way

I will only make it more clearer

x^2 >= 0 and y^2 >= 0 thus x^2 + y^2 >= 0

the sign "=" will happen when and only when all the sign "=" happen in the first two inequalities, thus x^2 + y^2 = 0 <=> x = y = 0

- Anonymous1 decade ago
1) x^2 + y^2 = 0

2) x^2 = -(y^2)

3) +/- x = +/- y * i (where i designates an imaginary number)

4) x = +/- y * i

case 1:

x1 = -(y*i)

case 2:

x2 = y*i

Treating x1 and x2 as complex numbers: we know that two complex numbers are equal iff their real and imaginary parts are equal...

The real part of the LHS of case 1 is x1, where as the real part for the RHS of case 1 is 0, therefore x1 = 0.

The imaginary part of the RHS of case 1 is -y, where as the imaginary part of the LHS of case 1 is 0, therefore y = 0.

Identical logic can be used for case 2, showing that the equation in step one is true iff x = y = 0.

- Anonymous1 decade ago
you would need one positive outcome and negative outcome to achieve a result of 0. but anything squared is a positive (eg. -3 squared is +9, not -9.) so x and y must both be zero, and a positive as a negative outcome cannot be added together to get 0, in this equation.

- PearlsawmeLv 71 decade ago
Let x^2 + y^2 = a^2.

This is the equation of a circle with radius a. (x , y) is any point on the circle.

If a = 0, then radius is zero and we have a circle which is a point (0,0). the only point on the circle.