Paul Erdős asked a question about this list. 1, a 3rd, a 7th, a 15th, and so on. To get the next fraction, double the bottom number and add 1. The top stays 1, and repeat that rule without stopping. Choose infinitely many different entries and add them. Must the total always be irrational, meaning it cannot equal a fraction of two whole numbers.

This is Erdős problem 257, and the general question is still open. I'll explain what this infinite sum means, what is known, and a complete proof of one of those results. Timestamps are in the description if you want to skip to a specific section.

We can write that rule using powers of 2, so choose a positive whole number, 1 or more, and call it n. 2 to the n means n twos multiplied together. n is called the exponent. So take away 1, then put 1 over that result. The top is 1, the bottom is 2 to the n minus 1. For exponent three, three twos multiplied to 8. Take away 1 to get 7, and the fraction is a seventh.

Exponent 1 gives 1. Exponent 2 gives a 3rd, and exponent 4 gives a 15th. Doubling 7 and adding 1 gives 15, as our rule says. We choose infinitely many different positive exponents, and put them in increasing order, and add their fractions. Each exponent is used once, so we can leave numbers out, but the choices do not stop. Choosing the exponent is the same as choosing entries from that

list. Add the chosen fractions in order. After one term, two terms, three terms, we have running totals, also called partial sums. The infinite sum means the number those totals approach. We need to show that there is such a number. Start with a half. Add a quarter, then an 8th. Each new term, half the previous one, without stopping. The running totals are a half, 3/4, 7/8, and so on. The gap to 1 halves each time, so the totals get as close

to 1 as we like. That's what we mean when we say this infinite sum equals 1. Now compare Erdős's list with 1. A half, a quarter, an eighth, halving each time. Each original fraction is at most its comparison fraction. At exponent 2, a third is below a half. At exponent 3, a seventh is below a quarter. For exponent n, the comparison denominator is a half of 2 to the n. 2 to the n is at least 2. Taking away only one leaves at least half of it

So the original denominator is at least as big as the comparison denominator

One over it is therefore no larger than the comparison fraction. The comparison fractions add to 2. 1 plus the halving sum we just saw. Leaving entries out only makes a partial sum smaller, so whichever infinite choice we make, its running totals keep increasing,

but they do not pass two. They approach a finite total. Every finite sum of fractions is a fraction. A third is 5/15. So a third plus a 15th is 6/15 or 2/5. But stopping there doesn't tell us what happens when we keep adding forever. Even an infinite sum can equal a fraction. The half, quarter, 8th example equals 1. So the question is whether this particular way of making the denominator's forces an irrational total for every infinite choice from the list. The general answer remains unknown,

but there is a known result if we add a condition on the exponents. The extra condition uses a second sum, one over each chosen exponent, and I'll call it the condition sum.

If that sum has a finite total, then the original sum is irrational. These are two different sums made from the same chosen numbers. Choose 2, 4, 8, 16 and keep doubling. One over those exponents gives a half, a quarter, an 8th,

halving at each step with total 1. Our original sum starts a third, a 15th, 1/255 with a fraction for every later doubling exponent. At exponent 4, the condition sum has a quarter, while the original sum has a 15th. If we keep every exponent, the condition sum is 1 plus a half plus a third and so on. That has no finite total, though our original sum still does,

so this theorem gives no answer here. Erdős already stated this result in 1968. He gave a proof when no whole number bigger than one divides any two chosen exponents and said that restriction could be dropped without giving the details. The paper linked below writes out a proof

without that restriction. I'd like to preface by saying I'm not a mathematician. I built the research system and directed the work. AI did most of the research and drafting using the sources and tools that I gave it and the results I'll explain go back to Golomb and Erdős.

We'll first prove the doubling sum irrational, then extend the argument to every choice that meets the finite total condition. The proof will rule out every possible denominator without calculating the sum's value.

We'll use a restriction that every fraction has starting with two fifths from the first two terms of the doubling sum. Three lots of two fifths is six fifths. Taking away one means taking away five fifths and this last fifth is what's left

sum. On this number line, slide that same piece back by one whole. Its end lands at one fifth. For each product, we take away all the whole ones it contains. The left over is at least zero and less than one. For multipliers

one and two, the leftovers are two fifths and four fifths. Three gives the one fifth we just found. Four takes away one whole and leaves three fifths. Five takes away two whole ones and leaves zero.

Zero is allowed but nothing above zero lands below one fifth because what's left is always a whole number of fifths. A tenth fits in that gap but this calculation can never produce

it. Now suppose our whole sum were a fraction p over q where q is the positive whole number on the bottom not necessarily five. Multiply it by a whole number, take away a whole number and what's left is an integer over q. The top has no fractional part though it can be negative. If that's above zero, the top is at least one so the result is at least one over q. So our remainder above zero and below one over q would rule out that fraction

try to make such a remainder by multiplying our sum and taking away whole numbers. We want the first few terms to come out as whole numbers so we can take them away. 255 is the third denominator and it's also a multiple of three and of fifteen. So multiply every term by 255. The

first three products are 85, 17 and one, which makes the whole number 103. The next product is one over 257, ironically, and every later one is smaller still. Each denominator

divides the next because of this multiplication. 15 times 17 is 255. One less than 16 times one more than 16 is 16 squared minus one. That works for any number so each doubling keeps the old denominator as a factor. Call the multiplier exponent k so the multiplier is two to the k minus one. Take away each whole part then add the separate leftovers down the column.

Here k is eight so we call that total r of eight. Each leftover is below one but added up r can be more than one. We only need the total we subtracted to be a whole number. Only finitely many whole parts come off. Past exponent k the denominator is bigger than the multiplier so every later product is below one and has no whole part. There's always a later positive term so r is above zero. So 255 times our sum is 103 plus r of eight. We've taken away our whole number and kept r positive

We still need to show that r is small. After multiplying by 255 the leftovers have exponents 16, 32, 64 and so on. Call one of those exponents n.

To bound r we compare each of these with a bigger fraction that's easy to add. First replace the top 255 by 256. Then make the bottom smaller. Half of two to the n. Both changes make the fraction bigger. For the comparison only fill in every missing exponent from 16 up adding positive fractions can only increase the total

These bigger fractions start at one over 128 and halve each time. So they add to one over 64. So R of eight is below one over 64. That rules out denominators up to 64. A

bigger denominator has a smaller gap so we need R smaller still. At any doubling exponent k every term up to k comes out as a whole number and the next chosen exponent is twice k. The same comparison starts at two over two to the k then halves each time so it adds to four over two to the k.

So fix any proposed denominator q first. We can choose a different multiplier for each proposed denominator.

The doubling exponents keep growing so we can pick one with two to the k. More than four times q. Then r is above zero and below one over q. Even though a fraction with denominator q would leave at least one over q. So the doubling sum is irrational. We've ruled out every fraction without calculating its value

Golomb had already proved this example irrational in 1963. The doubling proof used the large gaps between consecutive exponents. Our theorem doesn't require doubling. It only requires that one over the chosen exponent adds to a finite total. The square exponents meet that condition but the same choice of multiplier can go wrong. The square exponents are one, four, nine, sixteen. Multiply each positive whole number by itself and keep doing that.

Their condition sum is one plus a quarter plus a ninth plus a sixteenth with one fraction for every square. After the first term group two fractions then four then eight and keep doubling the group size.

The groups start at the squares of two, four and eight. So their largest fractions are a quarter, a sixteenth and one sixty-fourth. Two times a quarter is a half. Four times a sixteenth is a quarter. Eight times one sixty-fourth is an eighth

Each new group has twice as many terms, and a largest term a quarter as big

So the group totals are bounded by a half, a quarter, an eighth and the bounds keep halving.

The condition sum has a finite total. For the square exponents, our original sum starts one, a fifteenth, then one over five hundred and eleven. Multiply by fifteen and the first two terms give fifteen and one. Multiply by 255 and those terms are still whole numbers. All the later denominators are bigger than either multiplier. So those products stay below one and nothing comes off.

The larger multiplier is seventeen times the smaller, so it leaves seventeen times as much. Making the multiplier larger isn't enough. Instead, we'll compare several attempts. Each column here uses one multiplier on every term and its total left over R is at the bottom. We can choose any positive multiplier exponent. It needn't be one of our chosen exponents

Eight wasn't a square. Add the attempts remainders and divide by how many attempts we took. That is their average.

If every remainder were at least one over q their average would be at least one over q. as well.

So an average below one over Q means at least one attempt leaves less than one over Q, and we don't have to find which one. We'll make the early rows zero. For the remaining rows, an average below one over the rows exponent would be useful because the total of those one over exponent bounds can be made small. We'll prove that bound for complete repeating patterns. Some rows won't finish their patterns,

so we'll need a second bound for those. Pause the squares and take a separate row with exponent two, so its denominator is three. Try multiplier exponents one, two, three, and four.

The products are one third, one seven thirds and five. Seven thirds is two plus one third. So we take away two.

The leftovers go a third, zero, a third, zero, and repeat. For exponent three over seven, they go one, three, and zero, sevenths, then repeat. Now take exponent six over 63. With multiplier exponent four, the leftover is fifteen sixty-thirds. With exponent 10, 1,023 over 63 is 16 plus the same fifteen sixty-thirds.

Going from multiplier exponent four to ten, add six. So the power of two is multiplied by 64.

The numerator changes from 15 to 1,023. Their difference is 63 times 16. Dividing by 63 therefore adds 16 whole ones, with the same fifteen sixty-thirds left over.

For any row exponent n, adding n to the multiplier exponent multiplies the old power of two by two to the n. The increase is the old power multiplied by one less than two to the n. That last factor is exactly the row's denominator. Dividing the increase by that denominator leaves the old power a whole number, so the leftover repeats. So the leftover depends only on the remainder when we divide k by the row exponent.

10 divided by six leaves four. call that exponent remainder r. The leftover is two to the r minus one over the row's denominator. Here r is four, so the leftover

is fifteen, sixty-thirds. When r is zero, the leftover is zero. Starting at exponent one, the tops are one, three, seven, fifteen, thirty-one, zero, and then they repeat. The six entries up to that zero are one cycle. Now add one sixty-third to each of these six entries. The tops are now two, four, eight, sixteen, thirty-two, and one, which add to sixty-three exactly one whole. Take the added pieces away, and the real entries add to less than one. Divide by six

entries and the average is below a sixth, even though one entry is almost a half. For any row exponent n, the bigger tops are the powers of two from one up, n of them. Double their sum and take away the original, the middle cancels, leaving two to the n minus one, which is the bottom

So the bigger entries always make one whole, and a complete cycle of the real leftovers averages below one over n.

That's where the reciprocals in our condition come from, each bounds a row's average over a complete cycle. Now we need to keep the early rows at zero. We want the early rows to give zero on every attempt, which means every multiplier exponent must be a multiple of all their exponents. So choose a common multiple and move along

in steps of that size. For the squares, stepping by four keeps rows one and four at zero. With a step of four, the example row over sixty-three gives fifteen, sixty-thirds, three sixty-thirds, zero, then repeats.

Those three entries average two twenty-firsts. We can compare them with the full six entry cycle to prove that their average is below a sixth. The multiplier exponents are four, eight, and twelve. Dividing them by six leaves exponent remainders four, two, and zero. Another three steps add twelve, which is a multiple of six, so the same remainders repeat. Now leave the real attempts where they are and sort a copy by remainder. Zero, two, four.

The copied values are zero, three sixty-thirds, and fifteen sixty-thirds.

In the full cycle, each copied value is the smaller of a pair, zero beside one, three beside seven, fifteen beside thirty-one. Copy each value once more, so each appears exactly twice. That doubles the total, and the number of entries, so the average stays the same. And every copy fits under its partner from the full cycle. So the three attempt average is at most the full cycle average, which is below a sixth.

To use the same comparison for other rows and steps, we need to know which exponent remainders their cycles contain. Six and four are both divisible by two, and no larger whole number divides both. Call two, their largest shared factor. Every multiplier exponent is a multiple of four, and subtracting whole sixes leaves an even remainder. The only possibilities below six are zero, two, and four

For a row exponent n, and any step, call their largest shared factor, g. Each exponent remainder is a multiple of g, because both the step and the row exponent are multiples of g. Below n, there are n divided by g possibilities.

In our example six divided by two gives the three remainders we found.

A repeated remainder means the extra steps add up to a multiple of the row exponent. Here, four times the number of extra steps must be divisible by six. Four sixths reduces to two thirds. The three cannot cancel against the two, so it must divide the number of steps. The first possible repeat is after three steps. In general, reduce the fraction step over row exponent by dividing both by g. The bottom is n divided by g, with no factor left to cancel against the top.

So a repeat takes a multiple of that many steps. At that count, the bottom cancels and the remainder repeats. The first n divided by g remainders are therefore different. There are only that many possibilities, so one cycle visits each once. For six and four, the shared factor was two, and we copied each value twice.

In general, copy each selected value g times. It is the smallest value in its group of g consecutive remainders, so every copy is at most its partner in the full cycle.

Copying multiplies both the total and the count by g, leaving the average unchanged. So a complete stepped cycle averages below one over n for any step cycle. We still need a bound for a cycle we haven't finished.

However many attempts we make, some chosen exponent is larger than every multiplier exponent we've reached, so that row hasn't even reached its first zero. So we need a bound for an unfinished cycle. Take one more example row with exponent 5 over 31, and three attempts stepping by two. Exponents 2, 4, and 6. The leftovers are 3, 15, and one thirty-firsts, which average nineteen ninety-thirds, more than a 5th.

For the bound, fill in the skipped exponents from 1 to 6. The leftovers are 1, 3, 7, 15, 0, and 1, all over 31. One full cycle of 5, then the first entry of the next. The full cycle averages below a 5th.

For the unfinished part, look only at the increasing positive entries. Adding the later larger entries cannot lower their average, so the average of a beginning is no bigger than the average of all four positive entries. For a row exponent of at least 2, there are n minus 1 positive entries, at least half of n, and their total is below 1. So their average is at most 2 over n.

The row with exponent one is all zeros so the bound holds there as well. Back in our example, the row with exponent five, one full cycle and one extra entry each average at most two fifths. So the six entries add to at most six times two fifths. Take away

the entries we added and the sum can only go down. So the three actual entries still add to at most six times two fifths. So

Now divide by the three actual attempts. Their average is at most 4 5ths. 4 5ths is much larger than the actual 19/93rd, but this bound works for any number of attempts, and gets smaller as the row exponent grows. We use three attempts and filled in six exponents because the step was 2. Call the number of attempts m. In general, the filled in list has m times the step entries. Each complete cycle and the unfinished beginning of the last one averages at most 2 over n.

So that larger list adds up to at most m times the step times two over n.

Remove the entries we added, then divide by m, the number of actual attempts. Removing entries cannot increase the sum, and m cancels from the bound, leaving twice the step over n. The step stays fixed, and this works for every number of attempts. We can choose how many attempts to take later without losing the bound.

Now we can add the row averages over those same attempts. Back to the squares, with four columns, each a whole attempt. Row 9 is smallest at exponent 12, but row 16 reaches zero at exponent 16, and taking those two cells would mix two different multipliers, so average every

row over the same four columns. Adding across or adding down uses the same cells, so the row averages add up to the average over the column totals. That holds for any finite number of rows. Keep those four columns fixed as we include more rows. Each column total approaches its full value. There are only four columns, so the average approaches the average of those four full totals. Adding the row averages gives the average of the whole remainders as long as every row uses the same attempts. We can now put the proof together.

We'll use the doubling exponents again to work through the four choices. Take a proposed denominator of 5

We'll keep the middle rows together below a tenth of the average and the far rows together below another.

First, cut the rows after exponent 16. 1 over the later exponents adds to 1/16, below 1/10. Choose step 16, a multiple of all four early exponents, so those rows stay zero.

The cutoff picks rows. The step picks columns, but the loose bound on everything past 16 is 32 times 1/16, which is 2. Moving the cutoff and step together keeps that bound at 2, so we need two separate cutoffs.

Fix the step now, and move only the far cutoff. Put a second cutoff after 512. 1 over the exponents, beyond it, adds to 1 over 512, and 32 times that is a 16th, below the second tenth. The five rows between the cutoffs have cycle lengths 2, 4, 8, 16, and 32 attempts. 32 attempts would finish them all. We take 64 to show that more attempts leave the far bound unchanged, with multiplier exponents up to 1,024

Each middle row averages below one over its exponent so together their averages add to less than thirty one over five hundred and twelve. Add the far bound which is thirty two over five hundred and twelve and the average of our whole attempts is below sixty three over five hundred and twelve which is under a fifth.

So at least one whole attempt is below a fifth. It's above zero because the chosen exponent 2048 comes after every multiplier exponent we used. That puts it inside the fifth gap so the denominator

can't be five. The order matters. Choose which early rows to remove, fix the step, choose the far cut off, and only then choose how many attempts to take. Take any infinite choice where one over the chosen exponent adds to a finite total, and fix a proposed denominator q.

Because the condition sum approaches a finite total, the part still missing gets as small as we need. Choose the first cut off, so one over the remaining exponents adds to less than one over twice, q.

The early exponents have a common multiple. Their products will do, so use it for the step and keep it fixed. Then choose the far cut off, so twice the step, times the total of one over the exponents beyond it, is below one over twice Q. Only finitely many middle rows remain, so take a common multiple of their cycle lengths as the number of attempts. The early rows give zero, and the middle and far rows each give less than one over twice Q,

So the average of the whole attempts is below one over q which means at least one whole attempt is below one over q as well.

Every attempt is positive, so that this remainder is too small for a fraction with a denominator Q.

So any infinite choice of different positive exponents, with one over them adding to a finite total, the sum is irrational. Let's test the limit of this method by keeping every positive exponent where the condition fails.

After the initial one, the conditioned sum has a half, then two times at least a quarter each, then four at least an eighth each doubling the number of terms and halving their lower bound in each new block. Every block adds at least a half, so the total has no finite ceiling. For every multiplier two to the K minus one, with K at least one, the later terms of our original sum alone keep R above a half.

Swap their bottoms for two to the N, which makes each fraction smaller. That smaller tail adds to one, minus one over two to the K, at least a half, so R never gets below a half, and this method can't make it small. But the full sum is still irrational

Erdős proved that in 1948 by a different argument even though our condition fails for this sum.

Erdős himself wrote that he didn't think the condition was very relevant and he expected irrationality whenever the gaps between consecutive chosen exponents tend to infinity. Whether every infinite choice gives an irrational sum is still open.

The formal proof we'll look at next keeps the finite total condition. Here's the theorem written in Lean, a language for writing statements and proofs, so a computer can check the logical steps. The AI written formal proof covers more bases. We can replace the two in our fractions with any whole number from two upwards, under the same condition on our chosen exponents. Comparator checks a supplied proof against the statement written separately. This saved result is from the 24th of September.

It lists eight theorems including this one whose proofs were accepted by Lean's own checker and by a second checker, nanoda

The formal conjectures catalog hosted by Google DeepMind lists this case as solved, links to the Lean proof in my repository and keeps the general question open. The Lean proof uses limits of averages. We've worked with cutoffs and individual attempts. Checking the formal proof doesn't check the steps in this video and we still have to compare its statement with the paper. I'd love

to hear what you couldn't follow in this explanation and the written explanation is linked below in the expository paper

You can have a look at my Maths Map video in order to see a little bit more about what's going on behind the scenes of how I created this and what my intentions are for this project.

I hope you found this at least somewhat interesting if you're into mathematics or if like mathematics visualizations and you can check out my other videos for all the other things I've been working on so thank you very much for your time and I hope to see you in the next video