let $u$ and $v$ be two strings. Is $(u.v)^R$ equals to $u^R.v^R$?

Note: The $R$ notation means reverse order and the $.(dot)$ notation means concatenation.

  • $\begingroup$ Have you tried a few strings? For example, $u=0$, $v=1$? $\endgroup$
    – John L.
    Feb 17, 2019 at 10:55
  • $\begingroup$ @Apass.Jack I have tried to prove it. But I couldn't.So I thought maybe they are not equal. $\endgroup$
    – siaVash
    Feb 17, 2019 at 11:03
  • $\begingroup$ Let me rephrase my suggestion. Have you checked that equality with a few concrete instances? For example, what are $(u.v)^R$ and $u^R.v^R$ when $u$ is the string 0 and $v$ is the string 1? $\endgroup$
    – John L.
    Feb 17, 2019 at 11:06

1 Answer 1


No, $(u\cdot v)^R\not=u^R\cdot v^R$ more often than not.

For example, if $u$ is the string $race$ and $v$ is the string $car$. Then $$(u\cdot v)^R=(racecar)^R=racecar$$ while $$u^R\cdot v^R=(race)^R(car)^R=ecarrac.$$

Here are a few related exercises. All variables stand for strings.

Exercise 1. If $u$ or $v$ is the empty word, then $(u\cdot v)^R=u^R\cdot v^R$.

Exercise 2. If $u$ or $v$ are words of length 1 such that $(u\cdot v)^R=u^R\cdot v^R$. Then $u=v$.

Exercise 3. If $u=u^R$ and $v=v^R$, can we guarantee $(u\cdot v)^R=u^R\cdot v^R$?

Exercise 4. Prove that $(u\cdot v)^R=v^R\cdot u^R$.

  • $\begingroup$ I have to use induction to prove it? $\endgroup$
    – siaVash
    Feb 17, 2019 at 14:05
  • $\begingroup$ Yes, mathematically speaking. It becomes clear and obvious once you have played with several examples. $\endgroup$
    – John L.
    Feb 17, 2019 at 14:08
  • 1
    $\begingroup$ Inequality "more often than not": to be precise we have $(uv)^R = u^Rv^R$ precisely when $u$ and $v$ are powers of the same word $\endgroup$ Feb 17, 2019 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.