How to show that $\min\{\alpha,\max\{\beta,\gamma\}\}=\max \{ \min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}$ without considering too many cases? Here $\alpha,\beta,\gamma\in \mathbb{N}.$

I could consider $\alpha\leq \beta\leq \gamma$ and then various permutations of this inequality, but I was wondering if there was a cleverer argument.

PS. To give some context, this problem comes from the following problem: Let $a,b,c\in \mathbb{N}$ then show that $$a\wedge (b \vee c)=(a\wedge b)\vee (a\wedge c) $$ where $a\wedge b=\gcd(a,b)$ and $a\vee b=\text{lcm}(a,b)$ for any $a,b\in \mathbb{N}.$ Using prime factorizations for $a,b$ and $c$ one can reduce the problem to showing the equality that I have mentioned above.

  • 4
    $\begingroup$ I wouldn't consider $6$ to be "too many cases", but you can cut down to $3$ cases because of the symmetry between $\beta$ and $\gamma$. $\endgroup$ – Andreas Blass Mar 24 '18 at 14:03
  • $\begingroup$ You can cut it down to two cases: Either $\alpha\ge \max(\beta,\gamma)$ or $\alpha<\max(\beta,\gamma)$ $\endgroup$ – Mastrem Mar 24 '18 at 14:11

Suppose that $\alpha\ge\max(\beta,\gamma)$, so $\alpha\ge\beta$ and $\alpha\ge\gamma$, meaning that $$\max(\min(\alpha,\beta),\min(\alpha,\gamma))=\max(\beta,\gamma)=\min(\alpha,\max(\beta,\gamma))$$ Otherwise, $\alpha<\max(\beta,\gamma)$, so: $$\max(\min(\alpha,\beta),\min(\alpha,\gamma))=\max(\alpha,\min(\alpha,\min(\beta,\gamma)))=\max(\alpha,\min(\alpha,\beta,\gamma))=\alpha=\min(\alpha,\max(\beta,\gamma))$$


First case: $\alpha=\max\{\beta,\gamma\}$ then $$\min\{\alpha,\max\{\beta,\gamma\}\}=\alpha$$ and $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\beta,\gamma\}=\alpha$$

Second case: $\alpha>\max\{\beta,\gamma\}$ then $\alpha>\beta$ and $\alpha>\gamma$. These imply $$\min\{\alpha,\max\{\beta,\gamma\}\}=\max\{\beta,\gamma\}$$ and $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\beta,\gamma\}$$

Third case (last one): $\alpha<\max\{\beta,\gamma\}$ then $\alpha<\beta$ or $\alpha<\gamma$. Therefore $$\min\{\alpha,\max\{\beta,\gamma\}\}=\alpha$$ and $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\alpha,\min\{\alpha,\gamma\}\}=\alpha$$ or $$\max\{\min\{\alpha,\beta\},\min\{\alpha,\gamma\}\}=\max\{\min\{\alpha,\beta\},\alpha\}=\alpha$$ In either case the answers equal.


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.