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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.

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  • $\begingroup$ How about $1=2$? $\endgroup$ – DMcMor Apr 15 at 12:48
  • $\begingroup$ @DMcMor. I'm looking for formulas involving at least one variable. $\endgroup$ – Eleonore Saint James Apr 15 at 12:53
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    $\begingroup$ Then how about $a=a+1$, Eleonore? $\endgroup$ – Gerry Myerson Apr 15 at 12:54
  • $\begingroup$ Every arithmetical identity, like e.g. $(x-1)(x+1)=x^2-1$ holds for every value of $x$. $\endgroup$ – Mauro ALLEGRANZA Apr 15 at 12:57
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    $\begingroup$ Or $x>x$, or $x \cdot \tfrac{1}{x} = 0$. $\endgroup$ – Timon Knigge Apr 15 at 13:01
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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 $0\ne1$.

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