There are 719 errors in this sentence

Here’s a famous paradox (or a variant of it at least):

• There are two errers in this sentence.

The only visible error is the misspelt “errers”. But if the sentence claims to have two errors while having only one, that is another error and there are two errors after all.

Now for another variant. I’m not sure if I’ve thought this up for myself, but try this sentence:

• There are three errors in this sentence.

There are no visible errors in the sentence. Therefore it has one error: the claim that it has three errors when there is in fact no error. But if it has one error, it’s in error to claim that it has three errors. Therefore the sentence has two errors. And if it has two errors, again it’s in error, because it claims to have three errors while having only two. Therefore it has three errors after all.

The same reasoning can be applied to any integral number of errors:

• There are five errors in this sentence.
• There are 719 errors in this sentence.
• There are 1,000,000 errors in this sentence.
• There are 1,000,000,000 errors in this sentence.

No matter how large the number of errors, the sentence becomes true instantly, because each time the sentence makes a false claim, it makes another error. But those “times of error” don’t take place in time, any more than this equation does:

• 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16…

So I think these sentences are instantly true:

• There are infinitely many errors in this sentence.
• There are ∞ errors in this sentence.

But there are infinitely many infinities. Ordinary infinity, the infinity of 1,2,3…, is called ℵ0 or aleph-zero. It’s a countable infinity. Above that comes ℵ1, an uncountable infinity. So does this sentence instantly become true?

• There are ℵ1 errors in this sentence.

I’m not sure. But I think I can argue for the validity of sentences claiming fractional or irrational number of errors:

• There is 1.5 errors in this sentence.
• There are π errors in this sentence.

Let’s have a look at “There is 1.5 errors in this sentence”. There are no visible errors, so there’s one error: the claim that sentence contains 1.5 errors. So now there seems to be another error: the sentence has one error but claims to have 1.5 errors. But does it therefore have two errors? No, because if it has two errors, it’s still in error and has three errors. And that generates another error and another and another, and so on for ever. The sentence becomes unstoppably and infinitely false.

So let’s go back to the point at which the sentence contains one error. Now, the difference between 1 error and 1.5 errors is small — less than a full error. So how big is the error of claiming to have 1.5 errors when having 1 error? Well, it’s obviously 0.5 of an error. So the sentence contains 1.5 errors after all.

Now for “There are π errors in this sentence”. There are no visible errors, so there’s one error: the claim that the sentence contains π errors. Therefore it contains one error. But it claims to have π errors, so it has another error. And if it has 2 errors and claims to have π errors, it has another and third error. But if it has three errors and claims to have π error, it’s still in error. But only slightly — it’s now committing a small amount of an error. How much? It can only be 0.14159265… of an error. Therefore it’s committing 3.14159265… = π errors and is a true sentence.

Now try:

• There is -1 error in this sentence.

What is a negative error? A truth. So I think that sentence is valid too. But I can’t think of how to use i, or the square root of -1, in a sentence like that.

Pi and By

Here’s √2 in base 2:

√2 = 1.01101010000010011110... (base=2)

And in base 3:

√2 = 1.10201122122200121221... (base=3)

And in bases 4, 5, 6, 7, 8, 9 and 10:

√2 = 1.12220021321212133303... (b=4)
√2 = 1.20134202041300003420... (b=5)
√2 = 1.22524531420552332143... (b=6)
√2 = 1.26203454521123261061... (b=7)
√2 = 1.32404746317716746220... (b=8)
√2 = 1.36485805578615303608... (b=9)
√2 = 1.41421356237309504880... (b=10)

And here’s π in the same bases:

π = 11.00100100001111110110... (b=2)
π = 10.01021101222201021100... (b=3)
π = 03.02100333122220202011... (b=4)
π = 03.03232214303343241124... (b=5)
π = 03.05033005141512410523... (b=6)
π = 03.06636514320361341102... (b=7)
π = 03.11037552421026430215... (b=8)
π = 03.12418812407442788645... (b=9)
π = 03.14159265358979323846... (b=10)

Mathematicians know that in all standard bases, the digits of √2 and π go on for ever, without falling into any regular pattern. These numbers aren’t merely irrational but transcedental. But are they also normal? That is, in each base b, do the digits 0 to [b-1] occur with the same frequency 1/b? (In general, a sequence of length l will occur in a normal number with frequency 1/(b^l).) In base 2, are there as many 1s as 0s in the digits of √2 and π? In base 3, are there as many 2s as 1s and 0s? And so on.

It’s a simple question, but so far it’s proved impossible to answer. Another question starts very simple but quickly gets very difficult. Here are the answers so far at the Online Encyclopedia of Integer Sequences (OEIS):

2, 572, 8410815, 59609420837337474 – A049364

The sequence is defined as the “Smallest number that is digitally balanced in all bases 2, 3, … n”. In base 2, the number 2 is 10, which has one 1 and one 0. In bases 2 and 3, 572 = 1000111100 and 210012, respectively. 1000111100 has five 1s and five 0s; 210012 has two 2s, two 1s and two 0s. Here are the numbers of A049364 in the necessary bases:

10 (n=2)
1000111100, 210012 (n=572)
100000000101011010111111, 120211022110200, 200011122333 (n=8410815)
11010011110001100111001111010010010001101011100110000010, 101201112000102222102011202221201100, 3103301213033102101223212002, 1000001111222333324244344 (n=59609420837337474)

But what number, a(6), satisfies the definition for bases 2, 3, 4, 5 and 6? According to the notes at the OEIS, a(6) > 5^434. That means finding a(6) is way beyond the power of present-day computers. But I assume a quantum computer could crack it. And maybe someone will come up with a short-cut or even an algorithm that supplies a(b) for any base b. Either way, I think we’ll get there, π and by.