A question about Transcendental numbers

[ShayanJ]

hi pals
assume that m is a Transcendental number.
as you know we have:
x-m=0 => x=m
x^2-m=0 => x=+ or - sqrt(m)
.
.
.
so there must not be even one Transcendental number.
because all complex and real numbers can be the zero(s) of the equations above.
where is the problem?
thanks

 

[yyat]

A http://en.wikipedia.org/wiki/Transcendental_number" is, by definition, not the zero of a polynomial with rational coefficients.

 

[ShayanJ]

isn't 1 rational ?

 

[yyat]

isn't 1 rational ?

The constant term of the polynomial, in you case m, is also considered a coefficient (of x^0).

 

[HallsofIvy]

hi pals
assume that m is a Transcendental number.
as you know we have:
x-m=0 => x=m
x^2-m=0 => x=+ or - sqrt(m)
.
.
.
so there must not be even one Transcendental number.
because all complex and real numbers can be the zero(s) of the equations above.
where is the problem?
thanks

Do you understand the definition of "transcendental number"? A transcendental number is one that cannot be found as a solution to a polynomial equation with integer coefficients. Saying that "x= + or - sqrt{m}" says nothing about whether x satisfies a polynomial equation with integer coefficient.

 

[ShayanJ]

oh people really excuse me
i understand it now.
sorry for interrupting you