Square root of pi

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

that 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

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:

[tex]\sqrt{\pi} = 2 \int_0^{\infty} e^{-{x^2}} dx[/tex]

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

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:

[tex]a_n x^n + a_{n-1}x^{n-1}+...+a_1 x + a_0 = 0[/itex]

where [itex]a_i[/itex] are intergers.

matt grime

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

jcsd

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

NSX

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

jcsd

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

[tex]x^2 - 2\pi x + \pi^2 = 0[/tex]

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

fourier jr

yeah the polynomial has to have integer coefficients