So I'm trying to brush up on my algebra skills that I haven't used for 4 years and I've stumbled across a problem I'm having trouble comprehending.
Here is the question:
From what I can understand, you're reading it as 2 + 3/2 etc, while everyone else is reading is at 2^(3/2). Assuming that's what was meant, 2^(5/2) = root 32 = 2*root 8, 2^(3/2) = root 8, so 2^(5/2) - 2^(3/2) = root 8 = 2^(3/2). Hope that helps.
ohhh, dooohhhh. Now I think I got it. Like you said: I was looking at it like a fraction the whole time. Still need to level up on this concept. (and everything from Jr. High school -> calc)
mesh control wrote:My parents laughed in my face when I told them what I wanted to study.
That's really shitty of them. I'm a mathematics major, so I can help out too. But I'm guessing a lot of what you're doing is computation which I haven't done in a while so the physics/engineers may be better suited for that.
Three mathematicians walk into a bar. The waiter comes to their table and asks "Do all of you want beer?"
The first mathematician says "I don't know"
Then the second mathematician says "I don't know"
Finally the third mathematician says "Yes"
I understand what the person who came up with the joke had intended, but the reply of the third mathematician is incorrect. It should also be "I don't know."
Three mathematicians walk into a bar. The waiter comes to their table and asks "Do all of you want beer?"
The first mathematician says "I don't know"
Then the second mathematician says "I don't know"
Finally the third mathematician says "Yes"
I understand what the person who came up with the joke had intended, but the reply of the third mathematician is incorrect. It should also be "I don't know."
You don't get it, sorry
mesh control wrote:My parents laughed in my face when I told them what I wanted to study. Hmm, I wonder why I never went to college after high school....
That makes me
I bet the various math tutoring websites will be a good place to go to get some practice in (I should do this as well).
Libraries (including any community colleges) and thrifts are usually good places to pick up fairly recent textbooks on the cheap, too (and books on physics in general).
Three mathematicians walk into a bar. The waiter comes to their table and asks "Do all of you want beer?"
The first mathematician says "I don't know"
Then the second mathematician says "I don't know"
Finally the third mathematician says "Yes"
I understand what the person who came up with the joke had intended, but the reply of the third mathematician is incorrect. It should also be "I don't know."
You don't get it, sorry
Ah, now I do. If one of those who said "I don't know" wouldn't have wanted beer, that person would have replied with "no" instead, because then he would have known that not all of them want beer.
However, if the second one would have answered no, the third one would have had to answer with no as well, and the waiter would not have known whether to bring one or two beers.
It seems this question comes up a lot; must be a lot of people writing the same test. Although in the link above, the guy asking the question didn't make it clear that the 5/2 and 3/2 were supposed to be exponents, and neither of the people answering thought to think he could simply be misreading the question.
Friendly wrote:
However, if the second one would have answered no, the third one would have had to answer with no as well, and the waiter would not have known whether to bring one or two beers.
If one of them had said no, the rest wouldn't have said anything.
-Math 1 wants beer. However, he doesnt know if the 2 others want beer, so he logically answers "I dont know".
-Math 2 deduces, from the answer of math 1, that math 1 wants beer (if not, math 1 would ahve answered "no"). However, he doesnt know if math 3 wants beer, so he logically answers "I dont know".
-Math 3 deduces, from the answer of math 1 and 2, that both want beer (same proof as earlier). Moreover, he wants beer too, so he logically answers "yes".
Is that how the joke is supposed to be understood ?
The conclusion is that the waitress question is badly formulated in the sense that some cases may lead to her not knowing how many beers she must bring, exemple :
-Case <Beer, Not Beer, Beer> will lead to answers "IDK,NO,NO"
-Case <Beer, Not Beer, Not Beer> will lead to answers "IDK,NO,NO"
Same answers, not the same number of beers, to talk dirty : bijectivity (one-to-one) lost.
To talk dirtier, there is another way to see that one-to-one is lost : on one hand, there are 2^3 possibilities of Beer wanting states. On the other hand, due to the way the question is asked, the answers of mathematicians may only come in the form of a decreasing sequence of IDK, YES, NO ; decreasing in the sense that IDK > YES, IDK > NO and YES = NO. Due to the "decreasing" part there are strictly less than 2^3 possibilites of answers ("IDK, YES, IDK" is forbidden for instance).
Sump up :
-Beer wanating states = 2^3
-Possibilities of answer < 2^3
Conclusion : no bijectivity between them. Ask your question in a smartier way, you waitress !
Bravo jolie Ln, tu as trouvé : l'armée de l'air c'est là où on peut te tenir par la main.
You left out one possibility; if the first one wouldn't want beer saying "no", the others would have to reply "no" as well, which could mean that they want 0,1 or 2 beers total.
Now I read the written integration above again. Some notations misguide me, especially the x and the u. I assume this is simply because we europeans are not used to work with integrals this way, no problem.
But, finally, what do you prove ? That a primitive of e^x is e^x itself ?
Bravo jolie Ln, tu as trouvé : l'armée de l'air c'est là où on peut te tenir par la main.
guigui wrote:Now I read the written integration above again. Some notations misguide me, especially the x and the u. I assume this is simply because we europeans are not used to work with integrals this way, no problem.
But, finally, what do you prove ? That a primitive of e^x is e^x itself ?
guigui wrote:Let me put how I understand the joke :
-Math 1 wants beer. However, he doesnt know if the 2 others want beer, so he logically answers "I dont know".
-Math 2 deduces, from the answer of math 1, that math 1 wants beer (if not, math 1 would ahve answered "no"). However, he doesnt know if math 3 wants beer, so he logically answers "I dont know".
-Math 3 deduces, from the answer of math 1 and 2, that both want beer (same proof as earlier). Moreover, he wants beer too, so he logically answers "yes".
Is that how the joke is supposed to be understood ?
That is how I read it.
This is similar to a classical logic paradox, the "surprise execution." (Apparently this is called the "Unexpected hanging paradox.")
A man awaiting execution is told that he will be executed at noon one day next week, but it will be a surprise - he won't know in advance until the executioner arrives.
Working backwards, the prisoner realizes that he cannot be hanged on Friday, because if Thursday has passed, he would know that he has reached the only day left in the week available for the execution - but on Friday it will no longer be a surprise, since no other day is available.
Having eliminated Friday as a possible execution date, the prisoner realizes that he cannot be hanged on Thursday, because Friday would not be a surprise, but Thursday would also not be a surprise because he knows it cannot be on Friday, and now it can't be on Thursday either.
The prisoner decides that the judge is having a joke at his expense and that he actually will not be executed after all.
This reminds me of a scene from The Day of the Jackal, where a plotter against French President De Gaulle assures his lawyer after the last appeal fails: "You don't understand. No French soldier will raise his rifle against me." In the next scene he is shot by firing squad and a soldier with a revolver marches over to administer the coup de grace.
Three mathematicians walk into a bar. The waiter comes to their table and asks "Do all of you want beer?"
The first mathematician says "I don't know"
Then the second mathematician says "I don't know"
Finally the third mathematician says "Yes"
YESSSSSSS! My boss actually got it
Short answer for those still struggling. It's a logic thing. The bartender asked a logical (yes/no) question for the 3 mathematicians as a group, not as individuals. Any one of the 3 can violate the truth of the answer by saying NO. The first two mathematicians want a beer, so they don't say NO, but they also cannot answer for the remaining mathematician(s), so they can't answer YES. Only the last mathematician can confidently answer YES for the group, because the previous did not answer NO.
-ud
A chemist, physicist, mathematician and a computer programmer are asked the following question : "are all positive odd integers prime ?"
Chemist answer : "1 is prime, 3 is prime, 5 is prime, 7 is prime. Answer : YES".
Physicist answer : "1 is prime, 3 is prime, 5 is prime, 7 is prime, 9 is NOT prime, hum let's continue whatsoever, 11 is prime, 13 is prime. Answer : YES most of the time, we admit it as a fondamental rule"
Mathematician answer : "1 is NOT prime. Answer : NO".
Computer programmer answer : "1 is prime, 3 is prime, 5 is prime, 7 is prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime, 9 is not prime ..."
Bravo jolie Ln, tu as trouvé : l'armée de l'air c'est là où on peut te tenir par la main.
Three guests check into a hotel room. The clerk says the bill is $30, so each guest pays $10. Later the clerk realizes the bill should only be $25. To rectify this, he gives the bellhop $5 to return to the guests. On the way to the room, the bellhop realizes that he cannot divide the money equally. As the guests didn't know the total of the revised bill, the bellhop decides to just give each guest $1 and keep $2 for himself.
Now that each of the guests has been given $1 back, each has paid $9, bringing the total paid to $27. The bellhop has $2. If the guests originally handed over $30, what happened to the remaining $1?