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James

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Answer 1

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Let $N'=N+1$ and $K'=N+1$. Then
\[N' \geq K' \geq 2\]
so (1.2) holds for $N'$ and $K'$. Writing (1.3) for $N'$ and $K'$ instead of $N'$ and $K$ we get

\[ (1.3)' P_{K'}(K')< \dots P_{K'}(\frac{7K'-1}{2})=P_{K'}(\frac{7K'+1}{2})> \dots > P_{K'}(6K').\]
Substituting $N'=N+1$ and $K'=N+1$ we get
\[ (1.5) P_{N+1}(N+1)< \dots P_{N+1}(\frac{7(N+1)-1}{2})=\] \[P_{N+1}(\frac{7(N+1)+1}{2})> \dots > P_{N+1}(6(N+1)),\]
if $K'=N+1$ is odd.

Similarly substutitung $N'=N+1$ and $K'=N+1$ in (1.4) we get (1.6) if $N+1$ is even.

Erdos

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Erdos

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Leave a comment if you need any clarifications.
James

0

if u add 1 to both n and k, shouldn't u add 1 to 2?
Erdos

0

You are not adding 1. You are taking N' and K' to be your new N and K. You just want to make sure that $N'$ and $K'$ to be bigger than 2 so you can use the stated results.
James

0

and what's the reason behind making n' = k' it was N>= K >= 2
Erdos

0

If you just replace $N$ by $N+1$ and $K$ by $N+1$ in (1.3) and (1.4) we get (1.5) and (1.6). I will edit the answer in a minute.
Erdos

0

Done.
James

0

can u explain the reason why u made N' = K' >=2. It was initially N>= K >= 2
Erdos

0

N and K are just two general variables, and they can be replaced by any other variables.
Erdos

0

Also note that you still have N'>=K'>=2. We just know that N'=K' which also implies N'>=K'>=2.

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