Hey, /lit/. Please help me. I'm reading the tractatus logico

>I'm reading the tractatus logico philosophicus
don't

>>9973331

Too late now. I'm 50 pages in and I'm determined to finish it.

>>9973331
Can you just jump straight into Wittie's later stuff without going through this then?

>>9973326
the sum stands for considering all the possibilities in aggregate from the first (v) to the last(n)

>>9973326
I wonder why he put it that way. The sum, evaluated, just equals 2^n, doesn't it?

>>9973370
>implying witty knew hs-level math

>>9973370
because he's a pseud who got famous by dazzling other pseuds with entry level college math

>>9973354
yes

>>9973370
>>9973377
welcome to 20th century "philosophy"

>>9973370
>>9973376
>>9973377

Are the technical details in his shit really that silly? I mean, Russel strongly respected his insights, and he's clearly intelligent enough to overcome the flaws of analytic thinking, so this surprises me.

>>9973402
>Russel strongly respected his insights
Russel's opinions mean jack shit tbqh. Everything he said was either incredibly smart or incredibly stupid. It's hard to discern which is which

>>9973362

Hmmm, either it's too late here and I need some sleep or I genuinely misunderstood the last few sentences of the tractatus. Can someone elaborate a bit further on what Wittgenstein meant by the sum of the binomial coefficient?

>>9973626

oh nvm, I got it. I'm dumb. I'm going to bed.

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>>9973370

Yes. That is exactly right, and that is exactly the more pertinent observation.

OP, Witty is developing his own two-valued (true-false) logic in a particular way. By crude analogy, if you have these n light switches and each one can be either on or off at a given moment, then there are exactly 2 x 2 x ... x 2 (n times) = 2^n possibilities for the states in which the various light switches might exist - exactly because each light switch's state is independent of any of the others. That is effectively what the statement says, and it goes directly to truth-tables, truth-functions, etc, and a later statement about truth-functions which recycles the same notation for truth functions (4.42), this later statement is foundational of /how to build truth-tables/ with two-valued (they can be either true or false) propositions.

You should also especially pay attention to 5.101 which explicitly enumerates the sixteen possible truth-functions of two variables. Notice again that sixteen is a power of two. Two-valued true-false logic is central to the "meat of the math" of Wittgenstein's text.

But you're being good, and you want to /actually understand what the math says/. Another anon ITT correctly identifies it as low-tier undergrad/HS math, it just needs a bit of unpacking. You need to look up four or five different things and study them and /do exercises/ a bit: sigma notation (that's the specific part that you don't know), combinations, or the "choose" function (the thing in parentheses), mathematical induction, and above all PASCAL'S TRIANGLE, which shows all of this at-a-glance.

Look at pic related, and notice how the sums of each of the rows goes: 1, 2, 4, 8, 16, 32... A pattern. A pattern which can be proven with induction.

An exercise set for you to help you to better understand the Tractatus:

a) Find the definition of (n choose k) = nCk, where n and k are ordinary positive counting numbers, and convince yourself that the definition is true by thinking of examples.

b) Look up sigma notation on wikipedia and compute various sums using the notation. Notice especially that you do not have to compute infinite sums like the ones done in calculus, so don't worry about infinity just now ("infinite" sums are common in calculus. You're effectively just doing finite logic and philosophy at the moment so don't worry too much about the infinite).

c) Look up mathematical induction and learn how to prove certain things with it. This one involves the most growing pains.

d) /Prove/, using mathematical induction, that the formula in 4.27 is precisely equal to 2^n. Then you will be in a position to continue. (Hint which even some math people don't know: it is valid to "work the problem from both ends", knowing what you expect to end up with, and seeing if that agrees with what you start out with. If they don't join somewhere in the middle, then the thing is invalid. If they turn out to join, then your proof is valid and complete.

>>9973326
The capital sigma on the left is just a marker of the sigma-notation, which describes sums. To get the sum, you need to plug in for the variable on the bottom (in this case v) the number to the right of the equals sign (in this case, 0), into the formula on the right-hand side of the sigma. Calculate the result.

Now repeat this operation for every natural number following 0 (1, 2, 3...) up to n. So if n = 3, you need to plug in 0, 1, 2, and 3 into the right-hand side, and calculate the results for each.

Finally, you get the sum by adding up all these separate values you've calculated for each natural number.

So, for instance, n(SIGMA)v=0 2n, where n = 3, would result in 2(0) + 2(1) + 2(2) + 2(3) = 0 + 2 + 4 + 6 = 12.

So what does the thing on the right of the sigma mean? This is a notation for 'choose:' so the n over v in parenthesis means 'n choose v,' or nCv. This is also called a combination.

The idea behind a combination is: if you have a set of n things, how many unique ways can you pick out v things form that set? So for example, if you've got three balls, you might want to know how many possible combinations of two balls you can pick out of it. That's '3 choose 2.' Well, you can have three combinations: ball 1 and 2, ball 1 and 3, and ball 2 and 3. That's it – so 2C3 is just 3. There's a formula using factorials for calculating combinations, but it's not really relevant to this example.

So what Witty is saying is that if you have n possible states of affairs, you can figure out how many possible states the world can be in by calculating nCv for every number between 0 and n, and then adding all these up.

Say n = 2. Then this would be 2C0 + 2C1 + 2C2 = 1 + 2 + 1 = 4. So if there are two states of affairs, there are four possible ways the world can be. This makes sense: either both states of affairs holds, neither holds, the first but not the second holds, or the second but not the first does. As a helpful anon pointed out, this number will always be 2^n. So 3 states of affairs = 8 possibilities, and so on.

Hope this helps.

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>>9973647
>By crude analogy, if you have these n light switches and each one can be either on or off at a given moment

>implying light on/light off, visible/invisible isn't the constitutive metaphor for being/nothing in western-cartesian metaphysics
>implying your "crude" analogy isn't inheriting two thousand+ years of bad thinking

>>9973690
>That's it – so 2C3 is just 3
Sorry, this should read '3C2' is just 3. The above is my super-layman's way of trying to describe it, anyone with a better background in math feel free to correct it.

>>9973693
>excluded middle
>bad thinking