Show that non positive rational number, x, exceeds its cube, x^3 by exactly 1.
To prove that a non-positive rational number, x, exceeds its cube, x^3, by exactly 1, we need to demonstrate that x - x^3 = 1.
Let's start by assuming that x is a non-positive rational number. Since x is a rational number, it can be written in the form x = p/q, where p and q are integers and q ≠ 0.
Now, we can substitute this value of x into the equation and simplify it:
x - x^3 = (p/q) - (p/q)^3
To simplify further, let's find a common denominator for all the terms:
x - x^3 = (p/q) - (p^3/q^3)
Now, we can subtract the fractions:
x - x^3 = (p - p^3)/q^3
Since we want to show that x - x^3 = 1, we can equate the numerator to 1:
p - p^3 = q^3
Now, we have to show that there exist integers p and q such that p - p^3 = q^3.
One way to prove this is to demonstrate an example. Let's try p = 2 and q = 1:
p - p^3 = 2 - 2^3 = 2 - 8 = -6
q^3 = 1^3 = 1
As we can see, p - p^3 ≠ q^3 for p = 2 and q = 1. Therefore, we can conclude that there are no non-positive rational numbers x for which x - x^3 = 1.
Hence, we have shown that a non-positive rational number does not exceed its cube by exactly 1.