SOLUTION: Prove that cube root of 7 is an IRRATIONAL number.

Click here to see ALL problems on real-numbers

Question 91198: Prove that cube root of 7 is an IRRATIONAL number.
Found 3 solutions by stanbon, cparks1000000, ikleyn:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website!
Prove that cube root of 7 is an IRRATIONAL number ?
Assume that cube rt 7 is rational.
Then (7)^(1/3) = a/b where a and b are integers and a/b is reduced to lowest terms.
Then a=b[7^(1/3)]
Since a is a multiple of b and a is an integer, b divides a.
Since b divides a, a = nb and n is an integer.
Therefore 7^(1/3) = a/b = nb/b, so a/b is not reduced to lowest terms.
-----------
What led to this contradiction?
The assumption that 7^(1/3) was rational.
The assumption must be wrong.
Therefore 7^(1/3) if irrational.
================
Cheers,
Stan H.

Answer by cparks1000000(5)   (Show Source):
You can put this solution on YOUR website!
The previous solution is incorrect. Let me correct it:
Assume is rational. Then there exists integers and such that . Then by Zorn's lemma, there exists a greatest common divisor of and , say it is . Set such that and such that . Then by definition of the rationals, . But then divides . Then by definition of the rationals, . This is true since if we assume that divides ; then we can use the unique factorization theorem to conclude that where all of the are prime and none of them are 7. Thus divides which is a contradiction since it obviously has no factors which are since is prime. Thus we have that divides . Since divides , there exists such that , thus . Thus by similar logic to the above, we have divides . Since divides we can use the prime factorization theorem to once again conclude that divides . But then and must have a common factor. This is a contraction. Thus is irrational.

Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website!

Let us suppose that is a rational number. Then we can write = , where and are whole numbers. We can assume that the numbers and have no common divisor. Otherwise, we can cancel this common divisor in the numerator and denominator. Raise both sides of the equality = in the degree 3. You will get = . Hence, = . The right side of the last equality is divisible by . Hence, its left side is divisible by also. Since is divisible by , the whole number itself is divisible by . So, we can write = , where is another whole number. Now, substitute = into = . It gives = . Cancel both sides by . You will get = . The left side of the last equality is divisible by . Hence, its right side is divisible by also. Since is divisible by , the whole number itself is divisible by . We just got a contradiction. We assumed that = , where and are whole numbers with no common divisor, and the chain of arguments led us to the conclusion that both the integers and have the common divisor . This contradiction proves that the original assumption was wrong, regarding as a rational number. Hence, is irrational. The proof is completed. In the proof we used many times this property of the number 7: if 7 divides the product of two integers m and n, then 7 divides at least one of the integers. It is the common property of any prime number in the ring of integer numbers.