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For some equation to be an identity, should the LHS and RHS have same domain (given equality holds for all values of x). For example, are both of these an identity: $$1.\ \sec^2(\theta)-\tan^2(\theta)=1$$ $$2.\ \sec^2(\theta)=1+\tan^2(\theta)?$$

Are there any other conditions for the equation to be called an identity?

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  • $\begingroup$ Correct. See Identity (mathematics): "an identity between two expressions $A$ and $B$ (which might contain some variables) produce the same value for all values of the variables within a certain range of validity." Example: $(a+b)^2=(a^2+2ab+b^2)$ that holds for every value of $a,b$. $\endgroup$ Sep 25, 2020 at 13:26

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An equation is an identity when the left-hand side and the right-hand side agree at all points of the intersection of their domains. "Agree" means that arbitrary specialization of the variables appearing in the equation either produces an equality of values or is outside the domain of one or both sides.

If you additionally require that the domains of the two are equal, you are requiring equality of functions, which is a stricter requirement than identity.

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  • $\begingroup$ Are there any authoritative sources to confirm this statement? Does that mean that $\sqrt{-x^2} \equiv x$ is a valid identity on real numbers? The intersection of domains is {0} where LHS and RHS "agree". What about $\sqrt{-1 - x^2} \equiv 1.2345$? The intersection of domains is ∅, therefore LHS and RHS agree for all possible values of x (none). So we have then $\sqrt{-x^2} \equiv x$ and $\sqrt{-x^2} \equiv \frac{1}{x}$ but not $x \equiv \frac{1}{x}$, weird. I don't see any value in identities which are defined this way. $\endgroup$
    – il--ya
    Jun 5, 2023 at 11:03
  • $\begingroup$ @il--ya : Of sources that are careful in their definitions, some require that the intersection of domains is nonempty and some do not. There are all sorts of "stupid" identities. I think my favorite (?) stupid identity is (on the reals) $\frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arctanh}\, x = \frac{\mathrm{d}}{\mathrm{d}x} \mathrm{arccoth}\, x$ (vacuous because their domains, $|x| < 1$ and $|x| > 1$, respectively, have empty intersection, so there is no witness to failure of identity). $\endgroup$ Jun 7, 2023 at 21:02

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