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I was going through a paper where I stuck on a combinatorial argument as followsenter image description here

I want help with the first assertion i.e proving the inequality $\alpha(m+l)\le\alpha(m)$. As the author suggests it is very easy, I still could not figure this out although the rest of the argument was clear.

Kindly give your valuable suggestions thanks and regards in advance.

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Lucas theorem says that $$\binom{m}{\ell}\equiv \prod _{i=1}^N\binom{m_i}{\ell _i}\pmod 2,$$ where $m_i$ and $\ell _i$ are the bits on the binary decomposition of $m$ and $\ell$ respectively. This implies that $m_i\geq \ell _i$ for all $i$. Let $J=\{i: m_i=\ell _i =1\}$. If $J=\emptyset $, then $\ell = 0$ so you are done. If not, then either $i+1\in J$ or not: for the former, you would eliminate at least two zeros and recover one, so $\alpha (m+\ell)<\alpha (m)$. For the later you are replacing the one. Morally, if you get a bit $=0$, you recover at most one bit on the addition.

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  • $\begingroup$ Ca you explain"This implies that $m_i\geq \ell _i$ for all $i$" is it because of the given condition?@Phicar $\endgroup$ Commented May 15, 2022 at 13:58
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    $\begingroup$ @DevendraSinghRana Cause all the $\binom{m_i}{\ell _i}\equiv 1 \pmod 2$ by Lucas theorem. And so $m_i=0$ and $\ell _i =1$ can't happen cause the binomial is $=0$. $\endgroup$
    – Phicar
    Commented May 15, 2022 at 14:06

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