Show that 3√6is not a rational number
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Thanks for asking this question.
Let us suppose if possible 3√6 is a rational number and it is in simplest form of P/q as we did in last example √7 is not a rational number.
Then, p and q are integers having no common factor other than 1, and q ≠ 0
Now, p/q = 3√6 => (p/q)3 = 6 (i)
13 < 6 < 23 => 13 < (p/q) < 23 by using(i)
=> 1< p/q < 2
If q = 1, then p is an integer such that 1<p<2
But, there is no integer between 1 and 2.
Hence, q ≠ 1 and so q >1
(p/q)3 = 6 => p3/q3 = 6
=>P3/q = p3/q = 6q2 [multiplying both sides by q2]
Clearly, 6q2 is an integer (since q is an integer)
Since p and q have no common factor, p3 and q have no common factor.
Also, q > 1.
So, p3/q is not an integer, while 6q2 is an integer.
Consequently, p3/q ≠ 6q2.
Thus, we arrive at a contradiction.
Our supposition is wrong.
Hence, 3√6 is not a rational number or 3√6 ≠ Rational number.
I HOPE IT HELPS U.
HERE IS UR ANSWER.
Thanks for asking this question.
Let us suppose if possible 3√6 is a rational number and it is in simplest form of P/q as we did in last example √7 is not a rational number.
Then, p and q are integers having no common factor other than 1, and q ≠ 0
Now, p/q = 3√6 => (p/q)3 = 6 (i)
13 < 6 < 23 => 13 < (p/q) < 23 by using(i)
=> 1< p/q < 2
If q = 1, then p is an integer such that 1<p<2
But, there is no integer between 1 and 2.
Hence, q ≠ 1 and so q >1
(p/q)3 = 6 => p3/q3 = 6
=>P3/q = p3/q = 6q2 [multiplying both sides by q2]
Clearly, 6q2 is an integer (since q is an integer)
Since p and q have no common factor, p3 and q have no common factor.
Also, q > 1.
So, p3/q is not an integer, while 6q2 is an integer.
Consequently, p3/q ≠ 6q2.
Thus, we arrive at a contradiction.
Our supposition is wrong.
Hence, 3√6 is not a rational number or 3√6 ≠ Rational number.
I HOPE IT HELPS U.
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