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)); ```