Square Root Canal

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February 8th 2006, 07:09 PM
ThePerfectHacker

Square Root Canal

Here is one I came up myself. It is a proof.

Prove that $x^2=n$ has a unique positive solution in real number for $n\geq 0,n\in\mathbb{R}$ represented by $\sqrt{x}$ .

Please make in invisible letters.
February 16th 2006, 06:43 AM
TD!

Rather than only for the square root, you could prove this for $x^a = n$ .

A sketch of a possible proof: you use the completeness of the real numbers (axiom), i.e. you use the fact that every bounded set has a supremum (and, as follows, an infimum). You can then "sandwhich" $x^a \le n$ and $x^a \ge n$ from which follows that $x^a = n$ . You prove the uniqueness by contradiction, pick two and show they have to be the same.
February 16th 2006, 12:37 PM
ThePerfectHacker

Quote:

Originally Posted by TD!

Rather than only for the square root, you could prove this for $x^a = n$ .

A sketch of a possible proof: you use the completeness of the real numbers (axiom), i.e. you use the fact that every bounded set has a supremum (and, as follows, an infimum). You can then "sandwhich" $x^a \le n$ and $x^a \ge n$ from which follows that $x^a = n$ . You prove the uniqueness by contradiction, pick two and show they have to be the same.

The proof is simple, it is an application of the intermediate value theorem mostly.
February 16th 2006, 01:16 PM
TD!

There is no "the proof", unless you're referring to "your proof", it should be "a proof" ;)
March 3rd 2006, 05:42 AM
Rich B.

Elementary my dear Watson...quite.

Hi:

x^2 = n is equivalent to x^2 - n = 0. According to the Fundamental Theorem of Algebra, this polynomial has exactly two solutions. We know two solutions, both real and of opposite algebraic sign. Hence there is exactly one solution in R+. Whip the aforesaid into a rigorous proof and you're there.

Regards,

Rich B.
March 3rd 2006, 08:48 AM
ThePerfectHacker

Quote:

Originally Posted by Rich B.

Hi:

x^2 = n is equivalent to x^2 - n = 0. According to the Fundamental Theorem of Algebra, this polynomial has exactly two solutions. We know two solutions, both real and of opposite algebraic sign. Hence there is exactly one solution in R+. Whip the aforesaid into a rigorous proof and you're there.

Regards,

Rich B.

You made one mistake it has AT MOST two solutions and AT LEAST one, that is the fundamental theorem. Further, you need to show that the solution is real (which is easy).
March 3rd 2006, 03:39 PM
Rich B.

Greetings PHckr:

Okay, you can do that. But I did it without this theorem. Further, you made one mistake it has AT MOST two solutions and AT LEAST one, that is the fundamental theorem.

I am perplexed by the first two sentences noted above in blue print. I used one theorem, you used another (MVT). The FTA simplifies the proof with a single stroke of logic. Of course, such statement is meaningless should the remaining print in blue indeed be true. As it were, however, the fundamental theorem of algebra guarantees the existence of exactly n-complex roots for every polynomial of degree, n, having real coefficients. I believe it was Gauss (perhaps Euler --my memory for detail, sadly, varies inversely with age) who first conjectured existence of solutions numbering between one and n (this may be what you had in mind). However, the MVT, established more recently, asserts exactly n. Finally, as I state in my initial dialogue, yes, the plan suggested herein need only be formalized as a proof of, perhaps, three or so lines of text.

BTW, as an alternative to invoking the FTA, Descartes Rule of Signs affirms the singular positive root as well. What's more, such approach obviates any obligation to incorporate solutions into the formal proof.

Regards,

Rich B.
March 3rd 2006, 11:15 PM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

You made one mistake it has AT MOST two solutions and AT LEAST one, that is the fundamental theorem. Further, you need to show that the solution is real (which is easy).

The fundamental theorem of Algebra (in my version) states that every
polynomial equation of degree n greater than or equal to 1, with complex
coefficients, has a root in the complex plane.

Then a simple corollary shows that it has exactly n roots (counting multiplicity). in the complex plane.

RonL
March 4th 2006, 03:18 PM
ThePerfectHacker

Quote:

Originally Posted by CaptainBlack

The fundamental theorem of Algebra (in my version) states that every
polynomial equation of degree n greater than or equal to 1, with complex
coefficients, has a root in the complex plane.

Then a simple corollary shows that it has exactly n roots (counting multiplicity). in the complex plane.

RonL

I realized I was not clear enough in what I said. FTA states at least one. From here we can easily show (as CaptainBlack said) that there are at most $n$ solutions. But there are exactly $n$ linear factors.
March 4th 2006, 03:21 PM
ThePerfectHacker

Quote:

Originally Posted by Rich B.

Greetings PHckr:

Okay, you can do that. But I did it without this theorem. Further, you made one mistake it has AT MOST two solutions and AT LEAST one, that is the fundamental theorem.

I am perplexed by the first two sentences noted above in blue print. I used one theorem, you used another (MVT). The FTA simplifies the proof with a single stroke of logic. Of course, such statement is meaningless should the remaining print in blue indeed be true. As it were, however, the fundamental theorem of algebra guarantees the existence of exactly n-complex roots for every polynomial of degree, n, having real coefficients. I believe it was Gauss (perhaps Euler --my memory for detail, sadly, varies inversely with age) who first conjectured existence of solutions numbering between one and n (this may be what you had in mind). However, the MVT, established more recently, asserts exactly n. Finally, as I state in my initial dialogue, yes, the plan suggested herein need only be formalized as a proof of, perhaps, three or so lines of text.

BTW, as an alternative to invoking the FTA, Descartes Rule of Signs affirms the singular positive root as well. What's more, such approach obviates any obligation to incorporate solutions into the formal proof.

Regards,

Rich B.

What about $x^2+2x+1=0$ where is the second solution :eek: .
March 5th 2006, 03:52 AM
topsquark

Quote:

Originally Posted by ThePerfectHacker

What about $x^2+2x+1=0$ where is the second solution :eek: .

One solution is x=-1. The other is a complex solution: $x=e^{i \pi}$ . :D

-Dan
March 5th 2006, 05:49 AM
CaptainBlack

Quote:

Originally Posted by topsquark

One solution is x=-1. The other is a complex solution: $x=e^{i \pi}$ . :D

-Dan

But complex roots of real polynomials occur in conjugate pairs, so:

$x=e^{-i \pi}$

must also be a root giving three all told :p

RonL