Patrick B. answered 06/05/19
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Math and computer tutor/teacher
will prove by contradiction, for legs x and y .
suppose the diagonal is LESS than side x
sqrt(x^2+y^2) < x
x^2 + y^2 < x^2 <-- squares both sides
y^2 < 0
this is a contradiction: the square of a positive number must be zero or more.