Let k be a positive integer and p a prime number. Show that if p^2 is a factor of k, but p^3 is not a factor of k, this implies that k^(1/3) is not a rational number.
Regards
Do the coursework yourself you lazy bum.
as sam said
To show that k^(1/3) is not a rational number, we can use proof by contradiction.
Assume that k^(1/3) is, in fact, a rational number. This means that it can be expressed as a fraction in the form a/b, where a and b are integers that have no common factors other than 1, and b is not zero.
Since k^(1/3) = a/b, we can raise both sides to the power of 3 to eliminate the cube root:
k = (a/b)^3
k = (a^3)/(b^3)
Now, let's consider the prime factorization of k. We know that p^2 is a factor of k, but p^3 is not. Therefore, k can be expressed as:
k = p^2 * q
where q is some positive integer.
Substituting this expression for k in our previous equation, we get:
p^2 * q = (a^3)/(b^3)
Multiplying both sides by b^3, we have:
p^2 * q * b^3 = a^3
Since p and q are prime numbers, they can be considered co-prime, meaning they have no common factors. Similarly, a and b are also co-prime.
Now, let's consider the prime factorization of a^3. If p^2 is a factor of k, then it must also be a factor of a^3. Therefore, we can express a^3 as:
a^3 = p^2 * r
where r is another positive integer.
Substituting this back into our equation, we have:
p^2 * q * b^3 = p^2 * r
Notice that p^2 appears on both sides of the equation. We can cancel it out:
q * b^3 = r
Now, let's consider the prime factorization of r. Since r = q * b^3, we can conclude that p^2 is also a factor of r.
However, this contradicts our initial assumption that p^3 is not a factor of k. If p^3 were not a factor of k, then it shouldn't be a factor of r either. Thus, we have reached a contradiction.
Therefore, our assumption that k^(1/3) is a rational number must be false. Hence, k^(1/3) is not a rational number.
QED.