The following is a swift and easy proof that 1=0 in Maple. The problem is, that these situations might actually occur in real life computations.

#Let alpha be a primitive 10th root of unity
alias(alpha=RootOf(X^4-X^3+X^2-X+1,X));
#We consider a rational function in 2 variables, of degree 1 in both X and Y.
num:=X^2+Y-alpha^2;den:=(alpha^3+alpha)*(X-Y-alpha);
quot:=num/den;
#Now, it would seem OK to simplify this expression.
#Note that Maple insists on making the denominator have rational coefficients.
simpquot:=simplify(quot);
#So the numerator and the denominator of this simplified expression are:
num2:=numer(simpquot);den2:=denom(simpquot);
#Let's specialise (X,Y) to (alpha^3,0). Note that the original expression is
#defined here. it is not in the conjugate point (alpha,0), though.
x:=alpha^3;y:=0;
#so, num and den are both non-zero in (X,Y)=(x,y).
simplify(subs(X=x,Y=y,[num,den]));
#num2 and den2 (of the simplified expression) are zero, though.
simplify(subs(X=x,Y=y,[num2,den2]));
#so, the quotient can be evaluated with no problem in non-simplified form.
simplify(subs(X=x,Y=y,quot));
#but in simplified form, maple does not recognise 0/0.
simplify(subs(X=x,Y=y,simpquot));