Square Root of Pi: Is it Possible?

[Imparcticle]

The square root of pi is 1.777245... I know. But my math teacher says its impossible to determine a square root of an irrational number. Can anyone shed any light on this? is it or is it not possible to determine the square root (or any root) of an irrational number?

 

[matt grime]

you would have to explain what you mean by 'determine'

 

[Imparcticle]

I mean, Is it possible to arrive at a root of an irrational number?

 

[matt grime]

what is the "square root of pi", is still a vague question that needs elucidation. how on Earth would you "arrive" at root two? as root two squares to 2, surely if one can 'arrive' at root two, one can 'arrive' at root of root two since that is then the fourth root of 2, and if one can 'arrive' at square roots why not fourth roots.

 

[Integral]

Of course we cannot specify ALL digits of any irrational number, we can determine the number to pretty much any specifed precision.

 

[fourier jr]

The square root of Pi can't be constructed with straightedge & compass because Pi is transcendental. It's possible to construct (or "arrive at" ?) the square root of an irrational number that isn't transcendental though.

 

[Imparcticle]

So IOW, it is not possible to construct or "arrive at" a root for an irrational number?
not even, as Integral said, an approximate?

 

[arildno]

So IOW, it is not possible to construct or "arrive at" a root for an irrational number?
not even, as Integral said, an approximate?

You may definitely develop an algorithm that to arbitrary accuracy gives, for example, the decimal expansion of the square root of pi, as Integral said.
However, as Fourier jr. said, the particular technique known as "geometric construction" cannot be used to approximate transcendental numbers.

 

[matt grime]

there is nothing to stop you constructing the fourth root of 2 using compass and straightedge ie the square root of an irrational number, you might also be able to construct irrational roots of transcendental numbers as well. however did you even mean constructibility using ruler and compass?

of course the question springs to mind: how did you 'arrive' at or 'determine' the orginial irrational number that you wanted to find the root of?

i think you might mean "can we have an exact formula for the n'th position of the number in decimal notation" which of course makes the mistake of presuming that there is something special about decimal expansions, and that they are real numbers.

if you use continued fractions you'll find an 'exact' formula for many different irrational numbers.

 

[Zurtex]

You can obviously say that the square root of pi:

\sqrt{\pi} = 2 \int_0^{\infty} e^{-{x^2}} dx

It's all a matter of perspective, if you mean it can not be shown exactly in terms of decimal representation then neither can any other irrational number.

 

[HallsofIvy]

So IOW, it is not possible to construct or "arrive at" a root for an irrational number?
not even, as Integral said, an approximate?

On the contrary, Integral said it IS possible to approximate any number to any degree of accuracy. (Any real number can be approximated to any degree of accuracy by rational numbers- that's pretty much the definition of "real number".)

If, by "construct" you mean "construct, as in Euclidean geometry, with compasses and straightedge", then you CAN exactly construct any number that is "algebraic of order a power of two" and possible to construct such a number that is an approximation, to any degree of accuracy, of any number.

 

[jcsd]

Well pi squared is irrational, so if we can' 'determine' the root of an irrational number we can't 'determine' pi.

Perhaps your teacher was talking about transcendental numbers which are irrational (though no all transcendental numbers are irrational); pi is a transcendental number.

If a number is transcendental it is not a root to the equation:

a_n x^n + a_{n-1}x^{n-1}+...+a_1 x + a_0 = 0[/itex]<br /> <br /> where a_i are intergers.

 

[matt grime]

jcsd, about the bit in brackets: please show me a transcendantal (real) number that is rational then.

 

[jcsd]

jcsd, about the bit in brackets: please show me a transcendantal (real) number that is rational then.

Sorry, that was a typo it should of been 'not all irrational numbers are transcendental'

 

[NSX]

Well pi squared is irrational, so if we can' 'determine' the root of an irrational number we can't 'determine' pi.

Perhaps your teacher was talking about transcendental numbers which are irrational (though no all transcendental numbers are irrational); pi is a transcendental number.

If a number is transcendental it is not a root to the equation:

a_n x^n + a_{n-1}x^{n-1}+...+a_1 x + a_0 = 0[/itex]<br /> <br /> where a_i are intergers.

<br /> <br /> <br /> so you can&#039;t have something like (x-\pi)^2=0?

 

[jcsd]

so you can't have something like (x-\pi)^2=0?

No you can't have that as it specifies that the a_i terms must be intergers (though it's sufficent that the a_i terms are algebraic numbers to prove thatb the number is algerbraic) because if you expand that you get:

x^2 - 2\pi x + \pi^2 = 0

Therefore neither a_0 (i.e. \pi^2) nor a_1 (i.e. 2\pi) are intergers.

 

[fourier jr]

yeah the polynomial has to have integer coefficients

 

[HallsofIvy]

Of course, if a polynomial has all rational coefficients, then you could multiply through by the greatest common denominator to get integer coefficients so saying that an algebraic number is a number that satisfies some polynomial equation with rational coefficients is equivalent.

 

[NSX]

No you can't have that as it specifies that the a_i terms must be intergers (though it's sufficent that the a_i terms are algebraic numbers to prove thatb the number is algerbraic)

What do you mean by an algebraic number?

and what if the 2 roots two the quadratic can only be found using the quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

And let us [for this example] say that b^2-4ac is not a perfect square. Then the result of \sqrt{b^2-4ac} would create an irrational number (or not integer value)...

i.e. the 2 roots to the equation are \sqrt{2} and \sqrt{3}.

(x-\sqrt{2})(x-\sqrt{3}) = x^2-(\sqrt{3}+\sqrt{2})x + \sqrt{6}

 

[master_coda]

What do you mean by an algebraic number?

and what if the 2 roots two the quadratic can only be found using the quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

And let us [for this example] say that b^2-4ac is not a perfect square. Then the result of \sqrt{b^2-4ac} would create an irrational number (or not integer value)...

i.e. the 2 roots to the equation are \sqrt{2} and \sqrt{3}.

(x-\sqrt{2})(x-\sqrt{3}) = x^2-(\sqrt{3}+\sqrt{2})x + \sqrt{6}

But \sqrt{n} is a root of x^2-n=0. So as long as n is a rational number, we know that \sqrt{n} is algebraic.

 

[jcsd]

What do you mean by an algebraic number?

and what if the 2 roots two the quadratic can only be found using the quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

And let us [for this example] say that b^2-4ac is not a perfect square. Then the result of \sqrt{b^2-4ac} would create an irrational number (or not integer value)...

i.e. the 2 roots to the equation are \sqrt{2} and \sqrt{3}.

(x-\sqrt{2})(x-\sqrt{3}) = x^2-(\sqrt{3}+\sqrt{2})x + \sqrt{6}

It doesn't have to be a quadratic equation just a polynomial with integer coefficients.

For example \sqrt{2} + \sqrt{3} is algerbraic as it is a root to the equation:

x^4 - 10x^2 - 26 = 0

But there's no way of writing it as a polynomial of a lower order (e.g. a quadratic equation) with integer coefficients.

As Master Coda has shown both \sqrt{2} and \sqrt{3} are roots of quadratic equations anyway.

 

[NSX]

I see ... how does one go about proving such a thing?

I mean ... I could make up a function like f(x) = \sqrt{2}x^2 + \sqrt{3}x + 3

 

[matt grime]

it is a basic exercise that the roots of a polynomial with albegraic number coeffs are algebraic numbers (satisfy a polynomial with rational coeffs). one you should try.

incidentally you have not made up a function, that is an equation, and that is not the same thing.

 

[HallsofIvy]

You also did not make up an equation with INTEGER COEFFICIENTS.

A number is called "algebraic of order n" if and only if it satisfies a polynomial equation, of degree n, with integer coefficients, but no such polynomial of lower degree. As I pointed out before, if a number satisfies a polynomial equation with rational coefficients, then one can multiply through by the least common denominator of the coefficients to get an equation with integer coefficients. Because of that, the numberst that are "algebraic of degree 1" are precisely the rational numbers.