Question

you just found that c is \\( \sqrt{65} \\). is \\( \sqrt{65} \\) equal to a whole number? \\( a = 7 \\) \\( a ^ { 2 } + b ^ { 2 } = c ^ { 2 } \\) \\( 7 ^ { 2 } + 4 ^ { 2 } = c ^ { 2 } \\) \\( 49 + 16 = c ^ { 2 } \\) \\( 65 = c ^ { 2 } \\) \\( \sqrt { 65 } = c \\) \\( b = 4 \\) \\( c = ? \\) no, \\( \sqrt { 65 } \\) is not equal to a whole number. yes, \\( \sqrt { 65 } \\) is equal to a whole

Explanation:

Step1: Recall the definition of whole numbers

Whole numbers are non - negative integers: \(0,1,2,3,\cdots\)

Step2: Check the square of whole numbers around \(\sqrt{65}\)

\(8^2=64\) and \(9^2 = 81\)
Since \(64<65<81\), then \(8<\sqrt{65}<9\)

Answer:

No, \(\sqrt{65}\) is not equal to a whole number.