Need Math Help?

[moogle]

Yay, set theory. Doraki's got it. I'll just add a couple of definitions...

1) If you can prove that it's *possible* to have a one-to-one relation between two sets, then the sets have the same cardinality.
2) If you can prove that it's *impossible* to have a one-to-one relation between two sets, then the sets have different cardinalities.

And for those who are interested:
http://en.wikipedia.org/wiki/Countably_infinite#Gentle_introduction

 

[Haruka's DNA]

(Still AltF4 Posting):

Yea, Doraki's right. Galileo's paradox isn't much of a paradox. It's just something fun to think about. It's because you usually get 'tricked' into thinking of infinite sets of numbers as if they have a definite amount of elements.

Just like: What has more numbers, the set of all Real numbers from 0-10 or the set of all reals from 0-20? You might be tempted to say 0-20, but it's a trick question. They both have infinitely many elements, and you can't compare their "size" like that.

Oh, and the answer to the Infinite Hotel riddle:

The hotel manager comes up with an ingenious plan. He notifies everyone in the already booked Infinite Hotel to move up to the room that is twice their current room number. So the inhabitants of room 3 move to room 6. Room 6 moves to room 12. Now after having done this, all the odd numbered rooms are open! And we can fit yet another infinite number of people inside. Crisis averted.

 

[Jammer]

Thanks AltF4 and moogle and doraki. I totally understand (I saw what you guys were from the start, but I figured that if what you said was true, Galileo's paradox would be kind of stupid). So I guess Galileo's paradox is kind of stupid and obvious. I guess I should really learn this stuff in school before having discussions about it.

If I ever travel back in time and meet Galileo, and he tells me about his little paradox, I will look at him for a second, then say, "Duh. They're both infinite sets. Noob." In Italian so he understands me, of course.

I guess I'm a little disappointed.

Another disappointing paradox: If you move an element from one set to another set, you can raise the average of both sets. (The Will Rogers phenomenon.)

Once you realize how that happens, you think, duh. But until then, it's pretty interesting.

EDIT: Check it out! This is now the official geeky math thread! All Right!

 

[Lixivium]

Another disappointing paradox: If you move an element from one set to another set, you can raise the average of both sets. (The Will Rogers phenomenon.)

Once you realize how that happens, you think, duh. But until then, it's pretty interesting.

That's not a paradox at all, it's just a little less than obvious.

 

[MaNg0]

Wow this thread is way 2 complicated for me

ive been in algabrea 1 for 3 years =(

 

[Jammer]

Your avatar (Goofy) is especially apt for you, then, MaNgO.

Heh.

 

[cF=)]

sin²x + cos²x = -1

Discuss if algebra III will be a hard class or not if I had a good time in algebra II.

 

[Cobalt]

sin²x + cos²x = -1

I'm pretty sure it's actually 1, not -1...

1 - sin²x = cos²x, you know...

 

[cF=)]

Wrong you are, because circular trigonometry is not the same as hyperbolic trigonometry, hence why I talked about calculus 3.

 

[Cobalt]

Wrong you are, because circular trigonometry is not the same as hyperbolic trigonometry, hence why I talked about calculus 3.

Sorry, I was thrown off by the "Algebra II vs Algebra III" thing. In Algebra II/III, what I've written is correct.

 

[cF=)]

Yes, we are both right. Quantum theory allows that I guess...

lmao hivemind

 

[Best101]

Algebra is the easiest mess in the world to me. Don't bother asking me for help though, because I do go online very much.

 

[Keku]

uhh my teacher said to answer this question a couple years ago..concept it pretty easy, this was a question with a 20 dollar reward:

a/2 + b/3 + c/4=1

solve for a,b,and c

Uh? A = 0, B = 0, C = 4 for example.

 

[Colino]

Uh? A = 0, B = 0, C = 4 for example.

If I would get $20 for that in my 5 years of italian highschool I would have been the new bill gates

 

[AltF4]

sin²x + cos²x = -1

Discuss if algebra III will be a hard class or not if I had a good time in algebra II.

Do you mean to say:

sinh²x - cosh²x = -1

Uh? A = 0, B = 0, C = 4 for example.

I thought about the problem for a second, and realized that it's stupid.

If a,b, and c can be any integer, the problem is trivial. (Keku's answer)

So I thought maybe they had to be only positive integers. But then the problem is impossible!

The smallest sum you could get would be:

1/2 + 1/3 + 1/4 =
6/12 + 4/12 + 3/12 = 13/12 which is too big.

So in short, the problem is either given to us wrong, or is just stupid.

 

[cF=)]

Do you mean to say:

sinh²x - cosh²x = -1

/_ But, that's what I said!​

 

[AltF4]

Lol. I'm usually too serious with my posts. I should just go around making jokes more often. I think I've earned a little tenure.

 

[Keitaro]

Algebra is the easiest mess in the world to me. Don't bother asking me for help though, because I do go online very much.

Algebra was uber hard for me at least when you compare it to the Calc 3 I'm taking now. I think I just didn't have the motivation to do math back in highschool making me get every grade but an F throughout all 4 marking periods.

 

[Mr.GAW]

I think the teacher you have absolutely can help growth in math. While I've always been on the advance math course, some things took me over a week to completely understand when I was youger, but this year I have just an amazingly nice and relaxed teacher- and she helps me understand stuff so much quicker.

Just thought I'd share that.

 

[Jammer]

I think the teacher you have absolutely can help growth in math. While I've always been on the advance math course, some things took me over a week to completely understand when I was youger, but this year I have just an amazingly nice and relaxed teacher- and she helps me understand stuff so much quicker.

I'm happy you found such a good teacher.

I'm sure better teachers really help most people at "getting it" in math. I'm afraid to say that not a single one of my teachers has been better that another for me, though: They've all been pretty horrible.

I hate to brag (I really do), but I'm sure I'm not the only one in this thread with this problem. When you learn every single thing that's being taught in class the first time it's taught, and everyone else doesn't, and the teacher has to go over it again and again, well.... Math is the most boring class for me, even though it's the most interesting subject.

I've never had a single teacher who would teach me well. Most of them let me just sit there, playing games on my calculator so I wouldn't be disruptive. I'd look up every 5 minutes, see what's on the board, and go back to the game.

That's why I hate math class so much.

And by the way, it's still a problem with the school system, because I was in all the advance classes (in fact, doing advanced senior classes in my sophomore year). I ended up skipping 95% of the time in that class and still getting a 5 on the AP exam. And I still got in trouble for that.

You know what? Now I hate math. I avoid it whenever possible. And I think it's because it was just so fricking horrible to do the fricking homework every day and just sitting there, with nothing to challenge me or hold my interest. I blame the school system.

Sorry, just my little rant.

 

[Mr.GAW]

Okay, this is kind of embarressing- but I forgot something regarding difference/sum of cubes.

Let's use x^3-8, just as an example.

I know you have (x-2)(X^2+_x+4), but I forget how to find out what that middle figure is. I know that in this case it's 2, just because I've factored that so many times- but I don't really know how else to do it without actually multiplying those two out and seeing if it works with a certain number. Is it always just the opposite of the -2 in (x-2)? That would help a lot.

 

[moogle]

(x-y)(x²+xy+y²) = x³ + x²y + xy² - x²y - xy² - y³ = x³ - y³

(x+y)(x²-xy+y²) = x³ - x²y + xy² + x²y - xy² + y³ = x³ + y³

the answer to your question is yes.

 

[WFL]

Okay, this is kind of embarressing- but I forgot something regarding difference/sum of cubes.

Let's use x^3-8, just as an example.

I know you have (x-2)(X^2+_x+4), but I forget how to find out what that middle figure is. I know that in this case it's 2, just because I've factored that so many times- but I don't really know how else to do it without actually multiplying those two out and seeing if it works with a certain number. Is it always just the opposite of the -2 in (x-2)? That would help a lot.

For difference of two cubes I was given a formula that I never remember, but here is what I think it is. (a-b)(a^2+ab+b^2) in x^3-8, a=the cube root of x^3 and b=cube root of 8.

 

[TheFifthMan]

moogle, how do you do pretty print on the boards?

 

[Jammer]

He didn't do anything special. He just copied and pasted some special symbols.

Also, I answered a question in this thread! I'm so good at math...

 

[plasmawisp6633]

I have a challenge!!! I would have figured out this by myself, but I suck at counting principles and probability.

What is the probability that the last remaining tiles on a person's Scrabble holder are RCN (or NCR or CRN...)?

Here's the givens: 100 total scrabble pieces (that includes two blanks). There are 2 Cs, and 6 Ns and Rs. In this problem, there is no tiles on the other person's holder.

I appreciate the help.

*I normally don't bring back dead threads, but this was a special case*

 

[Doraki]

Assuming the player has the ability to form words with random letters with equal probabilities and is also fair in his choices, your problem is equivalent to computing the probability that when you pick 3 pieces randomly among the 100 pieces, you get 1 N, 1 C, and 1 R.

which is (2*6*6) * (3*2*1 / 100*99*98) = 432/970200 = 0.0445.. %

 

[AltF4]

Hey, my threads are never dead!


Imaging picking the pieces from the bag one-by-one.

The chances of you picking a C is 2/100, then after that there's a 6/99 chance to get an R, then a 6/98 chance to get a N.

That makes:

2/100 * 6/99 * 6/98 = 72/970200

But that's for the specific sequence "CRN". There are 3! = 6 ways to rearrange the letters, so you get:

6 * 24/970200 = 432/970200

That would be the chances of pulling out the letters RCN (in no particular order) from a bag as you described. Does that satisfy the problem?


EDIT: Woops. Doaki beat me to it! Plus I made a typo in my calculator. 2 * 6 * 6 is not 24.

 

[Jammer]

Is every even number the sum of two primes?

I need to have a proof either of this or its inverse by Friday, and I don't think I'll have enough time. Thanks for doing it for me, AltF4.

 

[AltF4]

Oooh, yea. That's a good one. Your teacher sure is mean.

That's a famous problem called "Goldbach's Conjecture". Long story short: it is as of yet unproven. It appears to be true, and people have calculated the numbers up to a great distance, but no proof actually exists.

So good luck with that.

 

[Jammer]

My teacher also said we had to find an odd perfect number, prove or disprove that there is an infinite number of primes whose value plus 2 is also a prime, find a Riesel number lower than 509,203 or disprove its existence, find a perfect cuboid or disprove its existence, and other things like that. For extra credit, we could divide a line into three equal parts using only a compass and a straight-edge.

If you ask me, I think she's trying to get the class to do her mathematician work for her.

And now I shall ask you, AltF4, to do these things for me, since you are so good at math. Thanks in advance.

 

[AltF4]

For extra credit, we could divide a line into three equal parts using only a compass and a straight-edge.

I know off-the-bat that that one's impossible. I remember that one specifically.

I wouldn't be surprised to see that all of these are unproven/unprovable.

 

[Jammer]

Hmm... It seems that either my jokes were unfunny, or you did not realize that I was joking.

You see, the compass and straight-edge one was extra credit because it was impossible. I don't actually have a teacher who said any of that.

It was mostly to freak you out, I guess. As in "Oh crap I can't do any of this!" I failed miserably in that respect, of course. You are one tough nut to crack, AltF4.

 

[AltF4]

Oh! Lol.

I'm sorry. I'm in the middle of studying for my finals, so I guess I was in a "problem solving mood". I didn't even think that you might be messing with me.

 

[Ripple]

complete the square help

-5(g^2)V(X^2) - 2gvX + 3(g^2)v = 0

given g and v are constants and X is a variable.

 

[digitalmaster287]

I was just wondering if there was a way to isolate for x in the following equation:

x sin x = a

Where a is a constant

My friend just asked me this and I had no idea what the answer was...

 

[Masque]

complete the square help

-5(g^2)V(X^2) - 2gvX + 3(g^2)v = 0

given g and v are constants and X is a variable.

Okay, this is a matter of mentally simplifying the situation. Since the quadratic term, --5 (g^2) v (X^2), is negative, factor that sign out right away. Keep the quadratic and linear terms inside the parentheses, but take the constant term outside (and disregard the sign change!), leaving the following:

- ( 5(g^2)v(X^2) + 2gvX + __ ) + 3(g^2)v = 0

My pre-calc teacher taught our class a little jingle to remember how to complete the square: DIVIDE BY TWO AND SQUARE IT! (Imagine the generic conga rhythm: da-da-da-da-daaaa-DA!) Since the linear term already has a coefficient of two, that makes our job a lot easier. Dividing the coefficients of the linear term and squaring them, we get (gv)^2. Add this INSIDE the parentheses, but take note of the negative sign outside the parentheses! Distributing the negative sign, you are technically subtracting the new (gv)^2 inside the parentheses, so balance it out by adding the quantity (gv)^2 outside. We now have:

- ( 5(g^2)v(X^2) + 2gvX + (gv)^2 ) + 3(g^2)v + (gv)^2 = 0

Kinda gross, yes. That's about as much as I can do for you, because I don't see a nice way to make this into two squares. :/ Sorry for being kinda useless. I can see that (gv) would be a term in the parentheses, but the 5 is throwing me off.

I was just wondering if there was a way to isolate for x in the following equation:

x sin x = a

Where a is a constant

My friend just asked me this and I had no idea what the answer was...

Hm... Were it just sin x = a, then you could use an inverse trig function to create x = sin^-1 a, or x = arcsin a. The x out front makes me leery, though, so I'm going to say that you can't isolate x in this situation.

 

[Gerudo Warrior]

you lost me at math ^^

 

[cF=)]

I was just wondering if there was a way to isolate for x in the following equation:

x sin x = a

Where a is a constant

My friend just asked me this and I had no idea what the answer was...

I don't remember but, couldn't derivative helps in this situation?

 

[Masque]

I don't remember but, couldn't derivative helps in this situation?

Well, x cos x + sin x = 0, and that won't really get you anywhere. But good thought, though!