Question 84483
The answer you were given to this problem by Checkley is WRONG!  YOU WERE CORRECT in saying that this is an irrational number.  An irrational number is a REAL NUMBER, but it is one that is NOT rational.  That is, it CANNOT be expressed as a quotient of two integers.  Irrational numbers include radical expressions that do not come out even, such as {{{sqrt(2)}}}, {{{sqrt(3) }}}, etc.  If the square root happens to be of a perfect square, such as {{{sqrt(25) }}}, then it is considered to be rational.  Your answer involves the square root of a number, 68, that is not a perfect square.  Therefore it is an irrational number. 


{{{sqrt(25/68)}}} is a REAL but yet an IRRATIONAL NUMBER.  


R^2 at SCC