If your favorite philosopher couldn't solve this then you've been following theories of some midwit pseud, congrats. What are some philosophers who could conceivably solve pic related? I imagine Descartes could have a shot at solving it, Whitehead and Russell probably too, any other non-brainlet philosophers I should read?
>>24797134Inb4 200 replies and no-one says anything the slightest bit consequential
>>24797137Fitting as philosophy has 3000 years of tradition of not saying anything slightly consequential lmfao
>>24797134I wouldn't know either. anyone have rec or general advice when studying math?
>>24797134Maths is not important to me, or to the philosopher.
>>24797134>show that 3 + 4 = 7 without adding 3 + 4a real philosopher would see that this is a retarded condition and that actual problems should be solved in any way available
>>24797134>lf your favorite philosopher couldn't Stopped there. If you have a favorite philosopher, then you’re a midwit.
>>24797170depends on your level of ability and what branch(es) of mathematics you intend to studyup through calculus I recommend the Art of Problem Solving series: comprehensive, logically presented, well-attestedbeyond that...probably ask /sci/
>>24797134The integer part of this sum is 26 because it's implied by the way you're posing the question, boom
>>24797134redditors really think that being a mathematician means grinding olympiad problems all day so that you can solve bullshit like this
>>24797189yeah, In the solution you do end up approximating the square roots, except you approximate them two at a time instead of one at a time.
>>24797289You could just draw squares and thus not have an approximation, it's a compass and straight edge problem
>>24797134The only number in the range which is a perfect root is 9.The other perfect roots are 4 and 16. Which are -1 and +1 away from 3. So,Root(5): 3 - e1, where e1 is something less than 1Root(6): 3 - e2, where e2 is something less than e1Root(7): 3 - e3, where e3 is something less than e2Root(8): 3 - e4, where e4 is something less than e3Then 3Now, Root(10): 3 + a1, where a1 is something more than 0Root(11): 3 + a2, where a2 is something more than a1Root(12): 3 + a3, where a3 is something more than a2Root(13): 3 + a4, where a4 is something more than a3The sum becomes: 3x9 + (- e1/2/3/4 + a1/2/3/4)As you can see, e1/2/3/4 are closer to 1, as the roots are closer to perfect square 4 than (3 + a1/2/3/4) are to perfect square 16. So the sum of -e1/2/3/4 is greater than +a1/2/3/4(Alternatively you can say that the graph of Root(x) is convex and differences decrease with increasing squares). You can integrate (3 - root(x)) between 5 and 8, and (3 + root(x)) between 10 and 13 to see that the sum value is between 0 and -1.So the overall quantity is 27 + some negative decimal value. Which proves that the integer quantity is 26.
This is all you have to do, it's compass and straight edge so no approximation
>>24797306If you drew the square roots with compass and straightedge then you would be calculating them exactly let alone approximating
/lit/ is so incredibly insecure about their lack of basic math skills, threads like these are always hilarious to read through
>>247971342 + 2 + 2 + 2 + 3.0 + 3 + 3 + 3 + 3 = 23There are 5 between 4 and 9. There are 6 between 9 and 16.The outputs of square roots from one whole number to the next increase at the same rate if one were to take all the numbers in between and treat them like ticks on a fuel guage. At 33%, for example, the numbers would be exactly the same. So, like:Between sqrt(1) and sqrt (4), there are 4 values inclusively speaking. Between sqrt(4) and sqrt(9) there are 6So1, 2(33.3...%), 3(66.6...%), 4.And4, 5(20%), 6(40%), 7(60%), 8(80%), 9And so on:9, 10(1/7th), 11(2/7th), 12(3/7th), 13(4/7th, 14(5/7th), 15(6/7th), 16And16, 17(11.111...%), 18(22.222..%), 19(33.333..%), 20(44.444), 21, 22, 23, 24(88.888...%), 25I still don't know what say the 20% tick would be as an actual number. 2.5 * 2.5 = 6.25, which would be the 45% tick. So, at 45%, the number should always be "(insert integer here).5" The sqrt of 20, should be very close. Indeed "4.4721..." .47 is very close to .5, so the rule works. To avoid approximation, there must be a rate by which we can take a percentage and turn it into a number, and then hypothetically we could simply add all those numbers together and see if the sum is in between 3 and 4 . I don't fucking know though. I'm not a mathmatician and want to get back to jerking it. This is too much work. There's only 8 numbers to worry about, 5 of which have remainders lower than .5, and 3 of which have remainders higher than .5 and lower than 1. But inherently what I'm doing now is approximation , so who gives a fuck. You know what the solution is? You draw all the sqaures, and then you cut them out with scizors, and then you draw a rectangle that's 1 by 1 by 27 and one that's 1 by 1 by 26. If you can cut up all the sqares and fit them into both rectangles, then there you go. You have your proof.
>>24797331It takes talent to explain things in such a way evven a brainlet like me understands.What is your age and ethnicity? Are you a jeet / chink or a White man?
A good scholars stationery should have a compass and straight edge
>>24797345this is the smartest post in the thread
>>24797356Thanks. Desu i thought my explanation was a little heavy-handed, but glad i could still get across.>Are you a jeet / chink or a White manOne of them.
>>24797331One correction "You can integrate (3 - root(x)) between 5 and 8, and *(root(x) - 3) between 10 and 13
>>24797331>As you can see, e1/2/3/4 are closer to 1, as the roots are closer to perfect square 4 than (3 + a1/2/3/4) are to perfect square 16. So the sum of -e1/2/3/4 is greater than +a1/2/3/4 But wait, how do we know that the sum isn't greater than 1? 27 - 1.1 is 25.9, meaning the integer part wouldn't be 26 in such a case.
>>24797331>You can integrate (3 - root(x)) between 5 and 8, and (3 + root(x)) between 10 and 13 to see that the sum value is between 0 and -1.Yeah, but how do you do that without approximating a square root?
>>24797404>>24797414Just write down the integrated equation in square root form, as I put it. You will be able to see that the final sum is less than 1. That's all you need.
>>24797423>You will be able to see that the final sum is less than 1. That's all you need.How? You end up gettin sqrt (2197) - sqrt (125) in your equation. How do you account dor this?
>>24797423Sophistry
ITTInsecure midwits trying to prove they're not midwits, as if anybody even gives a shit
>>24797436You've done some error
>>24797436Please see>>24797403
>>24797450What the hell are you talking about?
>That the lowest of all mental activities is arithmetic is proven by the fact that it is the only one that can also be executed by a machine, just as now in England this kind of calculating machine is already in frequent use for the sake of convenience. – Now however all investigation of the finite and the infinite at bottom amounts to a lot of reckoning. In this spirit we should evaluate the ‘mathematical profundity’ ridiculed by Lichtenberg when he says: ‘The so-called professional mathematicians, relying on the immaturity of the rest of mankind, have acquired a reputation for profundity that closely resembles that of holiness, which the theologians claim for themselves.’t. schopenhauerBonus:>we obtain only a few arbitrarily communicated results from them, and are in the same position as the man to whom the different effects of an ingenious machine are shown, while its inner connexion and mechanism are withheld from him. we are forced by the principle of contradiction to admit that everything demonstrated by euclid is so, but we do not get to know WHY it is so. we therefore have almost the uncomfortable feeling that we get after a conjuring trick, and in fact most of euclid's proofs are remarkably like such a trick. the truth almost always comes in by the back door, since it follows per accidens from some minor circumstance. frequently, an apagogic proof shuts all doors one after the other, and leaves open only one, through which merely for that reason we must now pass. often, as in the theorem of pythagoras, lines are drawn without our knowing why. it afterwards appears that they were traps, which shut unexpectedly and take prisoner the assent of the learner, who in astonishment has then to admit what remains wholly unintelligible to him in its inner connextion. this happens to such an extent that he can study the whole of euclid throughout without gaining real insight into the laws of spatial relations, but instead of these, he learns by heart only a few of their results... in our view, however, this method of euclid in mathematics can appear only as a very brilliant piece of perversity.
>>24797452First of all, the integral of sqrt(x) from 5 to 8 is not equal to sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8), so you could not use integrals to prove that the difference is less than one.
>>24797456Very good, just look at these people trying to work with squares while not even thinking about squares
>>24797460I knew you would say that. Just do this(3 - root(5)) - integral of (3 - root(x))[lim: 5 to 6]Do the same at other extreme 13 (but for (root(x) -3))Multiply each by 4 (as there are 4 discrete numbers, as a pair each side of root(9)). Now subtract them. Now you have the extra area jutting out from under the curve. Add to this to the previously calculated value. Now you mist agree this is the upper limit of (e1/2/3/4 - a1/2/3/4). And you will find it to be comfortably less than 1.
>>24797134sqrt(9) is 3. sqrt(8)+squt(10), sqrt(7)+sqrt(11), sqrt(6)+sqrt(12), sqrt(5)+sqrt(13) are all less than 6. The sum is less than 27. sqrt(x) is a concave function and sqrt(4) is 2 so it's obviously greater than 26.This is not an Olympiad question, a local KFC nigger can solve this, why ask a thing that every modern philosopher can solve?
>>24797191/thread
>>24797480I made a mistake
>>24797484You just wrote in broad strokes what I have already explained>>24797331
>>24797473Except how do you calculate any of this shit without approximating a square root?
>>24797134I don't think nick land would be able to solve this
>>24797552Just write the huge equations down with the values. You don't have to solve them. You will be left with rather straightforward integers and roots. Just check the upper bound and lower bound on the roots if required.Is saying root(3) is less than 2 but greater than 1 or 1.5 approximating it? Then you may as well give up on the question if you are gonna be caught up in semantics. By approximation they mean to prevent people from approximating each term to high accuracy then adding them, aka preventing brute force. You have to use some bounds to determine the final values.
>>24797588Pathetic
>>24797588Except you're adding and subtracting at least eight square roots, so there's no way you can tell if it's greater or less than one.Your math sucks and you should kill yourself.
>>24797610Write the equation.
>>24797134Well, its not that bad. We can just use some algebra and the an intuitive inequality. Sum = sqrt(5) + sqrt(6) + sqrt(7) + sqrt(8) + sqrt(9) + sqrt(10) + sqrt(11) + sqrt(12) + sqrt(13). Notice that we have 9 numbers, and that we have one integer in the list, sqrt(9)=3. We can arrange the numbers in pairs such that we take the numbers at each end and pair them up. Sum = [sqrt(5) + sqrt(13)] + [sqrt(6) + sqrt(12)] + [sqrt(7) + sqrt(11)] + [sqrt(8) + sqrt(10)] + sqrt(9) We can get a bound on each of this sums using the inequality: [sqrt(a) + sqrt(b)]/2 <= sqrt([a + b]/2). This only works because numbers get bigger as the argument of the sqrt gets bigger. so [sqrt(a) + sqrt(b)] <= 2sqrt([a+b]/2). You can verify that every pair just becomes 6. So we actually now just have 6 * 4 + 3, which is 27. So sum < 27. Getting a lower bound is more difficult, but you could just approximate the first decimal of each as a lower bound and take their sum. We'll get something like 26.d < sum < 27.
>>24797134Interesting problem. The way I looked at it was to sort of assume there would be a cute answer, which works better in exams than real life.The thing that jumps out is that the middle value (9) is a perfect square.So you have a perfect square, with four lower and four higher.So pairing them off is a natural thing to do.So we have the sum is:3 (the middle) +root 5 + root 13root 6 + root 12root 7 + root 11root 8 + root 10now you know from the shape of the log curve that these sums are going to be less than "averaging them out", i.e. each pair will be slightly less than 6, which is twice the middle value, which is 9, with root 3.So already we have an upper integral bound, which is 27. Total can't be that high, because we have 4 pairs, each pair slightly less than 6, and one 3 (the middle element).So now we just need to establish a lower integral bound.Again from the shape of the log curve, the closer the two elements get to each other, the nearer the sum gets to 6.So we consider root 5 + root 13. This is the lowest valued pair.We just need to establish that this is bigger than 5.75 and that will make our final sum bigger than 26 (because 4 * 5.75 + 3 = 26).Well root 5 is obviously bigger than 2.2 (22 * 22 = 484) and root 13 is obviously bigger than 3.6 (36 * 36 = 1296).So root 5 + root 13 is bigger than 2.2+3.6 = 5.8.So the overall sum is bigger than 26 and less than 27. QED.
You can tell people are retarded when they use QED instead of ὅπερ ἔδει δεῖξαι
>>24797634It's your equation. You write it down>noYou're full of shit.
>>24797679>you could just approximate the first decimal of eachDid you miss the part where it says Do not approximate the square roots
>>24797725>root 5 is obviously bigger than 2.2>root 13 is obviously bigger than 3.6 This counts as approximating the square root.
>>24797926Well it's only a lower bound. Not a lower AND upper bound. That doesn't count as approximating it as I understand the term.For example, is "bigger than 3" an approximation for pi? Not really.If there is some cute-as-a-button method of doing it without saying ANYTHING about the value of ANY individual root then OP will have to show us. Feels unlikely to me.
>>24797951Euclids spoke much about the alogon
>>24797134You can probably do this by allowing a variable level of precision and showing that the value converges to a within specific interval as the level of precision goes to infinite. Because the proof involves the precision going to infinite you are making no approximations, so long as you respect things like regions of convergence etc.Doesn't seem like all that special of a problem
These people are retarded
>>24797917You sound mad at that you couldn't figure it out
>>24798112No, you're just full of shit.
>>24798141Lol
>>24797306>>24797345>compass and straight edge>not approximating at every single step using precision of compass and straight edge
>>24797134My Wernicke's area is broken so I can only symbolize math in my head on occasion.
Hegel devotes a big chunk of his Science of Logic to problems like this.
>>24797134Tolstoy could solve it
>>24797134Only answers the "how" not the "why"
>>24797191>caring about the opinions of random anonsI shall do as I please.
>>24797456Who do you think had a higher IQ, Euler, or Schopenhauer? Keep in mind that Napoleon was much better at math than just about any other subject
The reason why people can answer questions like this is because they've been taught to solve similar problems before and were able to recognise that this problem is an example of X or Y
>>24798946but how did the teachers figure it out?? are they wizards???
>>24798143kill yourself
