Home >Opinion >Online-views >Opinion | Riemann—Standing on the shoulders of zeroes

Imagine a mathematician who has spent a couple of weeks asleep in a cave somewhere in, say, remote Telangana, somewhere where cellphone signals don’t reach. Sadly, she didn’t think of taking her 60-inch flat-screen TV with her. When she wakes up, she finds a way to check what’s been happening in the world outside. What would strike her as the biggest news story since she fell asleep, perhaps even over the last 150+ years?

Remembering she’s a mathematician, here’s my answer to that question: the Riemann Hypothesis has been proved.

I realize that’s underwhelming news to you; more than likely. But take my word for it, it would call for screaming headlines among mathematicians. In fact, mathematicians the world over have been in a whirl this week, because a professor emeritus at the University of Edinburgh, Michael Atiyah, has claimed he has a “simple proof" of the hypothesis. 

This is not a claim to take lightly. Atiyah is a past winner of both the Fields Medal (for mathematicians under 40) and the Abel Prize (the greatest honour in mathematics). The Riemann Hypothesis is—as Atiyah himself described it—“one of the most famous unsolved problems in mathematics and a formidable challenge" for him or anyone to prove. Put those together and you have the mathematical uproar we saw this week. 

I should point out that the Riemann Hypothesis is one of seven so-called “Millennium Problems" ( Solving any of them wins you a million dollars from the Clay Mathematical Institute. That much money is part of the reason for the buzz around Atiyah. But only part. The real story here is the sheer difficulty of proving the Riemann Hypothesis.

But of course, you probably want to know what the Riemann Hypothesis is. For that, bear with me through some mathematics. We start with the Riemann zeta function, always written with the Greek letter z (zeta), which is this infinite series:

z(s) = 1/1s + 1/2s + 1/3s + 1/4s

For example, let’s set “s" as 2.

z(2) = 1/12 + 1/22+ 1/32+ 1/42… 

That is,

z(2) = 1 + 1/4 + 1/9 + 1/16 …

This is a “convergent series". What that means is that even though it has an infinite number of terms, as you keep summing them you get closer and closer to a particular number, called the “limit". With s = 2, that limit is p2/6. So we would say z(2) = p2/6.

Setting s to any number greater than 1 (like 2) gives us a limit like the p2/6 we get for 2—the series will always converge. But things change if we use negative numbers for s. For example:

z(-1) = 1 + 2 + 3 + 4 … 


z(-2) = 1 + 4 + 9 + 16 …

Neither of those series converge. Instead, they diverge, or have no limit.

So far, nothing particularly interesting here, right? But things do start to get interesting if we use different kinds of numbers altogether for s. I mean complex numbers, those that involve the mysterious and imaginary “i", the square root of -1.

The thing about i is that it is indeed imaginary, in the sense that you cannot put a value to it, like 42, or 99.8. But if you accept that reality and then treat it like any other number, manipulate it like any other number, it turns out to be surprisingly useful. (In fact, that’s a good rule for various useful numbers—p, F (phi), e, i— that don’t look like the usual 1, 2, 3 and 42: treat them like any usual number.)

In any case, I won’t get into all that i is used for, here. But it serves to construct what we call complex numbers, like this: 2+3i. Or 9-7i. Just as you can think of the number 2+3, or 5, and manipulate that 5 any way you please, you can manipulate 2+3i any way you please too. Add 31? You get 33+3i. Multiply by 7? You get 14+21i. All you have to remember is that a complex number has a real part (like 14) and an imaginary part (like 21i).

Question: What happens if you use complex numbers in the zeta function?

Admittedly, this starts to get a little esoteric. How, for example, do we calculate the value of 22+3i and 32+3i and so on, which we will need to do to calculate the value of z(2+3i)? Actually, there is a well-defined way to do just that and, thus, to evaluate these series. I’ll spare you the mathematics, but I’ll say this: for any complex numbers whose real part is greater than 1—like 2+3i, or 5-43.5i—the series still converges. The limit is usually another complex number, yes, but it converges to a definite value.

In 1859, the German mathematician Bernhard Riemann showed that there is a way to extend the calculation of the value of the zeta function to complex numbers whose real part is 1 or less. That is, he showed that the zeta function has a definite value for any value of s—that’s right, including -1 (which we can write as the complex number -1+0i). We saw above that z(-1) is 1 + 2 + 3 + 4 … and concluded that it is divergent. But it was India’s own genius mathematician, Srinivasa Ramanujan, who showed, using Riemann’s methods, that z(-1) actually has the value -1/12. That is:

1 + 2 + 3 + 4 … = -1/12

I realize that’s a mind-bending result, but there’s serious mathematics behind it. (

So here’s the question Riemann tried to answer: what are the zeroes of the zeta function? That is, what are the values of s for which z(s) = 0? There are what he called “trivial" zeroes— the even negative numbers, like -2 or -842 (i.e. with no complex part). But what about non-trivial zeroes? 

Now we can show, and it is known, that all non-trivial zeroes of the zeta function have their real parts between 0 and 1. Which brings us finally to what Riemann hypothesized: that in all such zeroes, the real part is 1/2. 

Exactly 1/2. The Riemann Hypothesis for you.

If you’ll let all the earlier stuff about i and complex numbers flutter by, this is how simple the Hypothesis is—that the real part of the zeroes of the Riemann zeta function is exactly 1/2. And yet proving this has so confounded mathematicians for so long that it is now deservedly called the “most famous unsolved problem in mathematics".

Why, you ask, would anyone care about proving the Riemann Hypothesis? Well, Riemann also showed that his Hypothesis is intimately related to prime numbers. The further you get into the number line—the larger the numbers you’re looking at, that is—the rarer the primes become, though they never come to an end. Yet, even as they thin out, we can always predict how many primes we’ll find in a given stretch of numbers. Riemann produced a formula that calculates that figure and how primes are distributed, and the foundation of that formula lies in the zeroes of his zeta function. This is of great interest to mathematicians because primes themselves are truly the fundamental building blocks of mathematics. They are vital for you, too; for example, via their use in credit card transactions on the Web. (

No wonder mathematicians have tried long and hard to prove the Riemann Hypothesis. And this week, Michael Atiyah claimed he had done just that. He made available a five-page paper and gave a 45-minute lecture on Monday at the Heidelberg Laureates’ Forum, explaining his proof. 

Atiyah called it a “simple" proof, “even magical". It uses a well-known technique in logic, reductio ad absurdum: assume there is a zero whose real part is not 1/2, and show that this leads to a contradiction. On the way, Atiyah makes use of a “new function" named for his teacher J.A. Todd, speaks of “the transition from algebra to analysis", and mentions “the field of meromorphic functions". As you can imagine, magical or not, the mathematics is several leagues beyond me.

But more accomplished mathematicians are busy examining Atiyah’s effort. There remains considerable scepticism that he has really found a proof, but that’s the way mathematics goes. His result must be verified, and that process must necessarily start with scepticism.

But if I were you, I’d watch the news for the next few weeks. You might just run across the biggest news story since 1859. Mathematically speaking, at any rate.

Once a computer scientist, Dilip D’Souza now lives in Mumbai and writes for his dinners. His Twitter handle is @DeathEndsFun

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