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From: Dan Cass on 21 Jul 2010 09:42
> On Sat, 3 Jul 2010 22:07:49 -0700 (PDT), mike
> <mike_newsgroups(a)yahoo.com> wrote:
>
> >why isnt sqrt(x^2 + y^2) a polynomial?
>
> Assume it is a polynomial. By definition, there must
> be a finite
> expression with integer coefficients c_i_j of the
> form
> sum(over i and j) c_i_j * x^i * y^j
> which is the canonical expression for that
> polynomial.
>
> Evaluate the expression when x=1 and y=1. It is just
> the sum of the
> coefficients. Since they are all integers, the sum
> must be an
> integer.
>
> But the value of sqrt(1^2 + 1^2) is not an integer.
>
> Since we have reached a contradiction, our assumption
> must be faulty.
> Therefore, sqrt(x^2 + y^2) cannot be a polynomial.
>
> --
> Remove del for email

A nice proof, but the assumption you make:
> Assume it is a polynomial. By definition, there must
> be a finite
> expression with integer coefficients c_i_j ...
is necessary for your proof to work.
The coefficients c_i_j could presumably be rational
or even arbitrary reals.
From: Butch Malahide on 21 Jul 2010 14:03
On Jul 4, 8:08 am, Frederick Williams <frederick.willia...(a)tesco.net>
wrote:
> Axel Vogt wrote:
> > Frederick Williams wrote:
> > > Maybe the OP wants a proof that (x^2 + y^2)^(1/2) can't be put in the
> > > form P(x, y).
>
> > (x+y)^2 = x^2 + y^2 if char = 2, so P(x,y) = x+y will do.
>
> My guess is that the OP is interested in polys over R.

Let's see. Suppose sqrt(x^2 + y^2) is a polynomial in x and y. Setting
y = 0, the function f(x) = sqrt(x^2) = |x| is a polynomial in x. Since
f(x) vanishes at x = 0 but does not change sign, 0 is at least a
double root, i.e., f(x) is divisible by x^2. But the function f(x)/x^2
= |1/x| can't be polynomial with that infinite discontinuity at x = 0.
From: Axel Vogt on 21 Jul 2010 15:59
Butch Malahide wrote:
> On Jul 4, 8:08 am, Frederick Williams <frederick.willia...(a)tesco.net>
> wrote:
>> Axel Vogt wrote:
>>> Frederick Williams wrote:
>>>> Maybe the OP wants a proof that (x^2 + y^2)^(1/2) can't be put in the
>>>> form P(x, y).
>>> (x+y)^2 = x^2 + y^2 if char = 2, so P(x,y) = x+y will do.
>> My guess is that the OP is interested in polys over R.
>
> Let's see. Suppose sqrt(x^2 + y^2) is a polynomial in x and y. Setting
> y = 0, the function f(x) = sqrt(x^2) = |x| is a polynomial in x. Since
> f(x) vanishes at x = 0 but does not change sign, 0 is at least a
> double root, i.e., f(x) is divisible by x^2. But the function f(x)/x^2
> = |1/x| can't be polynomial with that infinite discontinuity at x = 0.

As an algebraic variant I think one can go the following route:

Assume t^2 - (x^2+y^2) = t^2 - f^2 = (t-f(x,y))*(t+f(x,y)),
which is the question in algebraic form.

Differentiating by x gives -2*x = -2*diff(f(x,y),x)*f(x,y)
and thus f(x,y) as a polynomial in x must have degree 1.
Dito w.r.t. y.

Therefore f(x,y)= a + b*x+c*y+d*x+y. Consider 0 = l.h.s - r.h.s.

Obviously a = 0 and then looking at y^2 one has c=0 or c=-2 and
after cancelling x:

0 = (-1+2*b*d+d^2+b^2)*x +- (2*b+2*d)*y = =((b+d)^2-1)*x +- (b+d)*y.

But "(b+d)^2-1 = 0 and b+d = 0" has no solution.
From: Ostap Bender on 22 Jul 2010 05:15
On Jul 4, 2:34 am, Frederick Williams <frederick.willia...(a)tesco.net>
wrote:
> Edson wrote:
>
> > Because, by definition, a polynomial in 2 variables is a function of the form
>
> > P(x, y) = Sum(i = 0, m) Sum(j = 0, n) c_i_j x^i y^j
>
> > f(x, y) = sqrt(x^2 + y^2) = (x^2 + y^2)^(1/2) is not of this form.
>
> Maybe the OP wants a proof that (x^2 + y^2)^(1/2) can't be put in the
> form P(x, y).

Maybe? That's CLEARLY what he wants: a proof that sqrt(x^2 + y^2)
cannot be expressed as a polynomial in x and y. Which is pretty easy
to prove by supposing that it is and then squaring both sides and
comparing terms.
From: Ostap Bender on 22 Jul 2010 05:19
On Jul 4, 12:26 am, Axel Vogt <&nore...(a)axelvogt.de> wrote:
> mike wrote:
> > why isnt sqrt(x^2 + y^2) a polynomial?
>
> The question is a bit ill-posed: what do you mean by 'sqrt'?
> About what 'objects' you are talking?

When people ask such questions, they are probably dealing with real
numbers, and 'sqrt' stands for 'square root'. Can you figure out what
the expression 'square root of a positive real number' means?
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