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Let me try to explain Schanuel's conjecture a little bit, if you're lost on words like transcendence degree. This turned out longer than I hoped so I hope it's not just a wall of text. sqrt(2), π, π/2, π^2, and e are all irrational. Now think about π and π/2: these are very closely related. What I mean is: π - 2(π/2) = 0 So we can add some combination of the two numbers, and get 0. Now what about π and e? These feel different. After working with them one gets the feeling that
aπ + be
will never be zero, if a and b are rational[1]. If that's true, then π and e are called "linearly independent over the rationals". We do not know whether π and e are linearly independent over the rationals. (In particular, we don't know whether e+pi is rational.) It turns out that stuff like that is hard. Now, there's an even stronger thing than being "linearly independent", and that's being "algebraicly independent". When talking about linear indepence we only thought about multiplying by rationals and adding (linear relations). For example, π and π/2 are NOT algebraicly independent. What about π and π^2? Well:
(π)^2 - (π^2) = 0
which is rational. This shows that π and π^2 are also not algebraicly independent. (The equation is obvious, but think about what it means: on the left of the minus, I have π, and I'm raising it to a power of 2. on the right of the minus, I have π^2, and I'm raising it to a power of 1).
We do not know whether π and e are mutually transcendental. So you should think: linear independence of rationals is a weak way of saying "these numbers aren't that related." Algebraic independence means there even less related. You can think of both as ways of proving huge amounts of numbers are irrational. If we can prove that e and pi are algebraicly independent then for example we would know
e + 5e^3 + e^7 + pi + 4pi^4
is irrational. Now, you can understand what this conjecture says. Take z_1, z_2, ..., z_n which are complex numbers. If you want, you can just imagine them to be irrational numbers. Suppose they are linearly independent over the rationals. The conjecture is that: at least n of z_1, z_2, ..., z_n, e^(z_1), e^(z_2), ..., e^(z_n) are algebraicly independent. [1] And not both 0 ;) Related: If you know some linear algebra, think about this. sqrt(2) is irrational. Think of 1 and sqrt(2) has "vectors" which you can multiply by scalars, but the only scalars you're allowed to multiply by are rational numbers. In this way you get a "vector space over the rationals" which includes things like 3+sqrt(2), 5, and 7sqrt(2). This vector space is also a field: we can multiply things together, and we get another element of the vector space:
(3+sqrt(2))(7sqrt(2)) = 3x7sqrt(2) + 7x2 = 14 + 21sqrt(2). Because sqrt(2) is algebraic over the rationals, this vector space-field thing is finite. But now do it with pi instead of sqrt(2), and you'll get an infinite dimensional vector space with the vectors 1, pi, pi^2, pi^3.... Elements of this field are things like 4+pi+7pi^3. |