Suppose not. Let p be prime and n = Sqrt[p]. Since p is an integer, if n is rational, then n is also an integer. So we have n.n = p. But since p is prime, only 1 and p divide p. -><- therefore n must not be rational
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.
2 is a prime number and its square root is an irrational number that cannot be expressed as a fraction
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.
No. The square root of an integer is either an integer, or an irrational number.
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.
The square root of the prime number 19 is an irrational number
Because 3 is a prime number and as such its square root is irrational
2 is a prime number and its square root is an irrational number that cannot be expressed as a fraction
The square root of 169 is 13 which is a rational and also a Prime number
√ 5 is an irrational number. Any square root of a prime number is irrational.
The square root of 121 is 11 which is a rational number and it is also a prime number
Because 7 is a prime number
17 is a prime number with no factors other than itself and 1 therefore minus square root of 17 is an irrational number.
No. 19 is a prime number, and all prime numbers have irrational roots.
'47' is a prime number, so it does not have a rational square root. sqrt(47) = 6.8556546... to 9 d.p. An irrational number which cannot be converted to a quotient; that is made into a fraction. NB The square roots of prime numbers are irrational.
No because the square root of any prime number is irrational inasmuch that it can't be expressed as a fraction.