Proving Real Numbers, Pythagorean Triples & Squares

[JanClaesen]

Can every real number be written as a sum of 2 real squares, and if not, how to prove that?
And how to prove that not every square of a whole number can be written as a sum of two whole squares?
I also read something that doesn't seem right to me: if a and b are rational, c must be irrational (counterexample: 3² + 4² = 5²), does the author mean a rational number that is not whole? And how to prove that?

I thank you.

 

[CRGreathouse]

Can every real number be written as a sum of 2 real squares, and if not, how to prove that?

i^2 = -1

And how to prove that not every square of a whole number can be written as a sum of two whole squares?

2^2 = ...?

 

[JanClaesen]

i^2 = -1

So... ?

2^2 = ...?

2^2 = 1^2 + 1^2... But that's not really what I meant with a proof.

 

[Tac-Tics]

So... ?

A square of a real is always non-negative. The sum of two non-negative numbers is non-negative. Therefore, negative numbers can't be expressed as the sum of squares.

 

[JanClaesen]

A square of a real is always non-negative. The sum of two non-negative numbers is non-negative. Therefore, negative numbers can't be expressed as the sum of squares.

I'm sorry, I didn't formulate my question quite well: I meant: can any (positive) real square be expressed as the sum of two squares?

 

[Tac-Tics]

I'm sorry, I didn't formulate my question quite well: I meant: can any (positive) real square be expressed as the sum of two squares?

Yes. For any positive real x, $$x = y^2 + 0^2$$ where $$y = \sqrt{x}$$.

 

[JanClaesen]

Yes. For any positive real x, $$x = y^2 + 0^2$$ where $$y = \sqrt{x}$$.

And if we exclude zero?

By the way, does anyone knows whether this is correct and how to prove it: if x and y are rational and not whole then z must be irrational.

 

[Tac-Tics]

And if we exclude zero?

Let's say $$x = y^2 + z^2$$ where both y and z are positive. One solution is let y = z, so we have $$x = y^2 + y^2 = 2y^2$$. Then $$y = z= \sqrt{\frac{x}{2}}$$ is a solution.

 

[CRGreathouse]

By the way, does anyone knows whether this is correct and how to prove it: if x and y are rational and not whole then z must be irrational.

If you mean "if x and y are rational but not integers then z = sqrt(x^2 + y^2) is irrational", it is not correct. Take (x, y, z) = (3/2, 2, 5/2).

 

[JanClaesen]

Thanks, any idea what the book was trying to tell, or what I'm trying to tell?
It was something Fermat had (really) proven (I'm not talking about his last theorem eh).

 

[ramsey2879]

Can every real number be written as a sum of 2 real squares

Simple. Say a is the square root of a positive number smaller than x

set $$b = \sqrt{x - a^2}$$ then $$a^{2}+b^{2} = x$$

 

[ramsey2879]

If you mean "if x and y are rational but not integers then z = sqrt(x^2 + y^2) is irrational", it is not correct. Take (x, y, z) = (3/2, 2, 5/2).

Not quite a counter-example since y is whole. Instead of dividing (3,4,5) by 2 divide (3,4,5) each by 5 or any whole number > 4.

 

[CRGreathouse]

Not quite a counter-example since y is whole. Instead of dividing (3,4,5) by 2 divide (3,4,5) each by 5 or any whole number > 4.

Fine. I took the statement as "when each of the numbers is an integer" and you took it as "when any of the numbers are integers". Either way examples are easy to find.

 

[Dadface]

X^2+[(x^2-1)/2]^2=[(x^2+1)/2]^2.Gives integral values for all odd values of x

 

[ramsey2879]

X^2+[(x^2-1)/2]^2=[(x^2+1)/2]^2.Gives integral values for all odd values of x

We were trying to make sense of "if a and b are rational, c must be irrational". The poster gave his own counterexample: 3² + 4² = 5², and questioned if the author meant by "rational" a rational number that is not whole? And how to prove that. Greathouse and I each gave essentially the same counter example since I based mine on his. My own feeling is that the original statement does not make sense because it is not given in the author's context. Surely, though, the poster was not looking for a response involving each of a,b and c in integer form as that was his own counterexample.