I'm given $g: (m,n) = 3^m 9^n, where (m,n)\in \mathbb{Z}^{+} \times \mathbb{Z}^{+}$ how do I prove this function is one-to-one?
All I've figured out so far is that
$g((m_1,n_1)) = g((m_2,n_2))$
$3^{m1} =3^{m2}$ and $9^{n1}=9^{n2}$ by dividing both sides by 3 and 9 $m_1= m_2$ and $n_1=n_2$ Therefore $g$ is one-to-one.

  • 1
    $\begingroup$ This function is not one-to-one. $\endgroup$ – nivekgnay Oct 6 '15 at 5:06
  • 1
    $\begingroup$ Isn't $g(3,1)=g(1,2)$? $\endgroup$ – user1337 Oct 6 '15 at 5:06
  • $\begingroup$ No wonder I find that there is something wrong with my prove. Thanks! $\endgroup$ – christinaqwer Oct 6 '15 at 5:07
  • $\begingroup$ Why $g(m_1,n_1)=g(m_2,n_2)$ implies $3^{m_1}=3^{m_2}$? (Apply that to the counter-example above) $\endgroup$ – Quang Hoang Oct 6 '15 at 6:11

As mentioned in the comments, the function is not one-to-one. For instance $g(2,0)=g(0,1)$.

The standard functions from $\mathbb{Z}^+\times \mathbb{Z}^+\rightarrow\mathbb{Z}^+$ that are one-to-one are given by $g(m,n)=p^mq^n$ for some distinct primes $p$ and $q$. Injectivity in this case follows from the uniqueness of prime decomposition.

  • $\begingroup$ Is this function also not an onto function? $\endgroup$ – christinaqwer Oct 6 '15 at 5:14
  • $\begingroup$ No, it never is. You can for instance not reach any prime different from $p$ and $q$ with this function. $\endgroup$ – This Is Me Oct 6 '15 at 5:34
  • $\begingroup$ How would I prove that it is not onto? $\endgroup$ – christinaqwer Oct 6 '15 at 6:10
  • $\begingroup$ Like I said, if you take a prime $p'$ different from $p$ and $q$, then by unique factorization into prime factors, you cannot write $p'$ as a product of powers of $p$ and $q$. For instance, for $g(m,n)=2^m3^n$ you cannot find $m,n$ such that $g(m,n)=5$ since 5 is not divisible by 2 or 3. $\endgroup$ – This Is Me Oct 6 '15 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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