# Example of mathematical antilogies involving the equality symbol ( always false statements, for all permissible values of the variables).

Logicians ( in propositional calculus) classify statements/formulas into 3 categories : tautologies ( always true) , contingent statements ( sometimes true, sometimes false) , antilogies ( always false).

I can find examples of mathematical "tautologies" , like (a+b)²=a²+b²+2ab.

I can find an example of mathematical contingent statement: a+a=a ( which is true if x=0, false otherwise), or a²=a ( true iff x=0, x=1)

But I cannot find an example of mathematical "antilogy" ( a statement that would be false for all permissible values of the variables) that would be an equality.

• How about $1=2$? – DMcMor Apr 15 at 12:48
• @DMcMor. I'm looking for formulas involving at least one variable. – Eleonore Saint James Apr 15 at 12:53
• Then how about $a=a+1$, Eleonore? – Gerry Myerson Apr 15 at 12:54
• Every arithmetical identity, like e.g. $(x-1)(x+1)=x^2-1$ holds for every value of $x$. – Mauro ALLEGRANZA Apr 15 at 12:57
• Or $x>x$, or $x \cdot \tfrac{1}{x} = 0$. – Timon Knigge Apr 15 at 13:01

Since this was received well in the comments, and since it's generally considered better to have answers posted as answers: $$a=a+1$$
Digression: Of course, a lot depends on the phrase, "permissible values of the variables". If only natural numbers are permissible values, then $$a+1=0$$ answers the question. If only integers are permissible, $$a+a=1$$. If only rationals are permissible, $$a^2=2$$. If only reals, $$a^2+1=0$$. One might even object that $$a=a+1$$ is not an antilogy, if infinite cardinals are permissible. So perhaps one has to go to $$a-a=1$$ for an example of a one-variable equation that is an antilogy in any theory in which subtraction is a binary operation and $$0\ne1$$.