How can I prove that a number is irrational?
Let r be a root of x^3 - 2x + 5 = x^3 - x^2 + 9. Show that none of r, r^2, or r^3 is irrational.
+18 To prove that a number is irrational, you need to show that it cannot be expressed as the quotient of two integers (i.e., it cannot be written in the form a/b, where a and b are integers and b is not equal to zero). One way to do this is to assume the number is rational and derive a contradiction.
To show that none of r, r^2, or r^3 is irrational, we can use a proof by contradiction. Assume that one of them (let's say r) is irrational. Then, we can write r as a/b, where a and b are integers and b is not equal to zero. We can also assume that a and b have no common factors, since if they did, we could simplify the fraction.
Now, let's substitute r into the equation x^3 - 2x + 5 = x^3 - x^2 + 9:
(r)^3 - 2(r) + 5 = (r)^3 - (r)^2 + 9
Simplifying this equation, we get:
(r)^3 - (r)^2 - 2(r) + 4 = 0
Multiplying both sides by b^3, we get:
a^3 - a^2b - 2ab^2 + 4b^3 = 0
This is a polynomial equation with integer coefficients. Since r is a root of this equation, a/b must also be a root. But we assumed that a and b have no common factors, so a^3 and b^3 have no common factors either. Therefore, by the rational root theorem, any rational root of this equation must be of the form ±p/q, where p is a factor of 4 and q is a factor of a^3.
However, we know that r is not rational, so it cannot be of this form. Therefore, we have a contradiction. This means that our assumption that r is irrational must be false, and r must be rational.
Using the same argument, we can show that r^2 and r^3 must also be rational. Therefore, none of r, r^2, or r^3 is irrational.
