If a/b=sqrt(6), then a2=6b2
On the other hand, given integers ''a'' and ''b'', because the valuation (i.e., highest power of 2 dividing a number) of 6b2 is odd, while the valuation of a2 is even, they must be distinct integers. Contradiction.
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An irrational number is a number that never ends. An example of an irrational square root would be the square root of 11.
The square root of 18 is an irrational number because it cannot be expressed as a fraction.
Nice question! The square root of (any number that isn't a perfect square) is irrational. No prime number is a perfect square. So the square root of any prime number is irrational.
No. The square root of an integer is either an integer, or an irrational number.
Yes. The square root of any number that is not a perfect square (like 9 or 36) is irrational.