»  Solutions to puzzles in my VDARE.com monthly Diary

  April 2022


The Math Corner of my April Diary included this brainteaser:

Brainteaser:  What is the first digit after the decimal point in the number you get by raising the square root of 3 plus the square root of 2 to the billionth power?


•  Solution

Ordinary calculators don't help. The one that comes with my Windows 10 gave me √3 plus √2 without complaining: it's 3.1462643699419723423291350657156 and change. The log of that to base ten likewise: 0.49779521211533912848434277312126…. Multiply that by a billion: 497,795,212.11533912848434277312126….

So the actual value of (√3 + √2) to the billionth power is 10497,795,212.11533912848434277312126…. Remembering that na+b = na×nb, that's equal to 10497,795,212 × 100.11533912848434277312126….

The latter number there is 1.3041847825658114556867010045569…. So the answer we have is a number with 497,795,213 digits to the left of the decimal point, those digits starting off with 13,041,847,825,658,114,556,867,…. There are a whole lot more digits — nearly half a billion of 'em — before we get to the decimal point and see what's after it.

Can the Windows 10 calculator give me all those nearly-half-a-billion digits? Er, no.

Time for some subterfuge. The fact of a billion being an even number, while we are dealing with square roots, is suggestive. Instead of a billion, let's try an even-number exponent the calculator can manage. Let's try ten. What is (√3 + √2)10 ?

The desk calculator can handle that OK. The answer is 95,049.999989479221461223705617444. How about twenty? Answer: 9,034,502,497.999999999889313218938. Wow! Look at those nines after the decimal point in both cases! How about an even number not divisible by ten? How about 26? Answer: 8,763,458,109,129.9999999999998858898.

So is the answer to my brainteaser nine? Is it the case that raising (√3 + √2) to some humongous even number will give you a big whole number minus some teeny teeny decimal quantity, leaving a bunch of nines at the right of the decimal point? And if that is so, why is it so?

Yes, it is so. For some positive integer n, expand (√3 + √2)2n using the binomial theorem. I'll use aCb to denote the ordinary binomial coefficients, which are of course all positive integers. I'll put those binomial coefficients in parentheses, just for reading clarity. Here we go.

    (√3 + √2)2n = 3n + (2nC1)3n-1√6 + 2(2nC2)3n-1 + 2(2nC3)3n-2√6 + 4(2nC4)3n-2 + 4(2nC5)3n-3√6 + 8(2nC6)3n-3 + 8(2nC7)3n-4√6 + 16(2nC8)3n-4 + 16(2nC9)3n-5√6 + …

There are 2n +1 terms on the right-hand side there. The 1st, 3rd, 5th, … terms are all positive whole numbers; the 2nd, 4th, 6th, … are all irrational.

That's convenient. It gives us an easy way to get rid of all the irrational terms. All we have to do is expand (√3 − √2)2n. That gives us the same expression but with all the even-numbered terms — all the irrational ones! — having negative signs. Ha! Add the two expressions together, those irrational terms all disappear. So

    (√3 + √2)2n + (√3 − √2)2n = (some large positive integer).

But (√3 − √2) is 0.31783724519578224472575761729617…, which is of course less than one. If you raise a number less than one to some integer power, you get a smaller number: the square of one-third is one-ninth, the twentieth power of 0.317837…  is 0.000000000110686…. Raising (√3 − √2) to some big integer power will give you a very tiny number indeed.

So our last expression now looks like this:

    (√3 + √2)2n + (some very tiny number way less than 1) = (some large positive integer).

Subtract that tiny number from both sides of the equation and … QED. That first digit after the decimal point is 9; so are several of the ones following it.

In fact (√3 − √2) is 10-0.49779521211533912848434277312126…. If you raise that to the billionth power you get 10-497,795,213 × 100.88466087151565722687873…. The latter term there is 7.6676251200587693018790353626249…; so the billionth power of (√3 − √2) is 0.0000…0000766762512…, where the number of zeros right after the decimal point is 497,795,212.

If you subtract that very tiny number from any positive integer N the result will be N − 1, followed by a decimal point, followed by 497,795,212 nines, followed by 23323748799412306981209646373750….

So, supererogatorily comprehensive solution to the brainteaser: the first digit after the decimal point is nine, and so are the next 497,795,211 digits.