Question

is sqt rational and transcedental

Answer 1

no if you mean the square root of an integer.

Answer 2

sqt3

Answer 3

\[\sqrt{3}\] is the solution to
\[x^2=3\] so no

Answer 4

the square root of any number is NEVER transcendental because it's always involved in the solution of some polynomial (namely the quadratic)

Answer 5

as before, transcendental means no the solution to a polynomial equation.

Answer 6

@jim
i was being careful because of course
\[\sqrt{\pi}\] is transcendental

Answer 7

ah true, you have a point

Answer 8

so i wanted to make sure the question meant the square root of a positive integer

Answer 9

*rational number

Answer 10

but you can easily say that \[\large \sqrt{\pi}=x\] which would mean that \[\large x^2-\pi=0\], which is a perfectly valid polynomial

Answer 11

hold on. by that argument all numbers are transcendental

Answer 12

how do you know if a number is transcendental

Answer 13

oh ok just looked it up again

Answer 14

@jim
root of a polynomial with rational coefficients.

Answer 15

the coefficients have to be rational, my bad

Answer 16

you could just as well say
\[\pi\] is transcendental as it is the solution to
\[x-\pi=0\]

Answer 17

yeah thx just remembered that lol

Answer 18

got scared for a moment!

Answer 19

lol totally overlooked the simple equation of x=pi --> x-pi = 0

Answer 20

almost got close to proving pi = 3..(again)...doh...