ChapterZero’s relocated

I still haven’t hammered out the database issues, but I’m tired of my little vacation (i.e. several months!) from posting, so I’m going ahead with the relocation. At some unspecified time in the future (once I figure out the technical details), I’ll migrate the database over from the old location. That site’ll still be up, albeit in its broken state (the math and other escaped code is not being interpreted correctly).

Here’s an inagural problem: Argue that \[\frac{21}{22} \geq \frac{\log 9}{\log 10} \geq \frac{20}{21}\] without using decimal expansions of the involved quantities. Feel free to use arguments that aren’t feasible without a CAS like Mathematica– I sure did.

  • Great to see that your blog is back up.
    Why the change? You’ve had ‘tangetspace’ for years.

    By the way, the blog entry says you posted this on July 28, but only tonight, 6 Aug, did you updated your old blog with direct link to this updated site.
    (As a matter of habit, I check your blog daily. This time was done at 3 in the morning.)

    The problem that you posed took me almost an hour.
    I must learn to stop doing math problems at 4.30 in the morning.

    I shall email you my solution so that others may have a crack the problem.

    • swiftset

      I’m switching to a cheaper service– I’m paying $40/mo for the old server, while the new one is less than $100/yr. This is something I should have done a long time ago.

      Yep, I forgot to post a pointer from the old blog to here. I’ll check out your solution 🙂

  • Welcome back, Alex and it seems you also switch to the beautiful MathJax for math displaying — awesome! I’ve tried several moving for my blog, but paused at the amount of work anticipated work to work out the database and stop…

    My two cents about the problem: it’s not hard to see the first half is equivalent to establish the ordering between $10^{21}$ and $9^{22}$, but
    (9 + 1)^{21} = \sum_{i=0}^{21} {21 \choose i} 9^{i}
    {21 \choose i} 9^i = 21\times 20 \cdots \times (21-i +1) 9^i > 9^{21} , \forall i \geq10.
    So $10^{21} > 9^{22}$, and I think the second half can be worked out in similar way.

    • Sorry Alex my argument the last displaying equation is incorrect… thinking of improvement

    • swiftset

      Yeah, moving an established blog is harder work than I’d imagined. I checked out your blog, you have some interesting stuff, so I’ll be following 🙂

  • Jeff

    Followed the same path as Ji Sun above. To show log9/log10 > 20/21 we try to show
    21*log9 > 20*log10 or 21*log9 > 20* log 9 *10/9 or 21*log9 > 20* log9 + 20*log10/9 or log 9 > 20 * log 10/9 . Squaring we have 9 > (10/9)^20 or 9^21 > 10^20. Now I have to try and use a binomial expansion of (1+9)^20 to compare it to 9^21. Here is where I get stuck. I was hoping to group terms and show that each one is less than 9^20. For instance, take the first three, then two for each of the next few terms. I don’t even know if this is true but it seems worth a shot.