If \( p \in\{0 ; 1 ; 2 ; 3 ; 4\} \), determine the value of p such that \( \sqrt{\frac{16}{p-2}} \) 1.1 .1 is irrational 1.1 .2 is non real 1.1 .3 is undefined

To determine the value of \( p \) such that \( \sqrt{\frac{16}{p-2}} \) is irrational, non-real, or undefined, we need to analyze the expression's denominator \( p-2 \). 1.1.1 For \( \sqrt{\frac{16}{p-2}} \) to be irrational, \( p-2 \) must be a perfect square. The only value for \( p \) from the set \(\{0, 1, 2, 3, 4\}\) that makes \( p-2 \) rational (and thus keeps the fraction under the square root rational) is \( p = 3 \), yielding \( \sqrt{16} = 4\) (not irrational). Hence, all other values, except \( p=2 \) yield a non-perfect square. 1.1.2 To have \( \sqrt{\frac{16}{p-2}} \) give a non-real number, the expression \( p-2 \) must be negative. In this case, \( p \) must be less than \( 2 \). This occurs when \( p = 0 \), \( p = 1\). Both cases yield a negative in the denominator resulting in complex numbers since a square root of a negative number is not defined in the reals. 1.1.3 The expression is undefined when the denominator is zero. Thus, \( p - 2 = 0 \) leads us to \( p = 2\), which makes this condition true, indicating an undefined state. In summary: - For irrational: \( p = 4 \) makes it rational but approaching irrational through larger sequences. - For non-real: \( p = 0\) or \( p = 1\). - For undefined: \( p = 2\).