## April 2022

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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?

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**• 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 10^{497,795,212.11533912848434277312126…}. Remembering thatn^{a+b}=n^{a}×n^{b}, that's equal to 10^{497,795,212}× 10^{0.11533912848434277312126…}.

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 calculatorcanmanage. 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 pointin 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,whyis it so?

Yes, it is so. For some positive integern, expand (√3 + √2)^{2n}using the binomial theorem. I'll use_{a}C_{b}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}= 3^{n}+ (_{2n}C_{1})3^{n-1}√6 + 2(_{2n}C_{2})3^{n-1}+ 2(_{2n}C_{3})3^{n-2}√6 + 4(_{2n}C_{4})3^{n-2}+ 4(_{2n}C_{5})3^{n-3}√6 + 8(_{2n}C_{6})3^{n-3}+ 8(_{2n}C_{7})3^{n-4}√6 + 16(_{2n}C_{8})3^{n-4}+ 16(_{2n}C_{9})3^{n-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! — havingnegativesigns. 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 asmallernumber: 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 numberwayless 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}× 10^{0.88466087151565722687873…}. 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 thatverytiny number from any positive integerNthe result will beN− 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.