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[Fossaman]

So, question for you folks who do a whole bunch of math at a whack: How do you set up your work area? I start to get a sore neck and headache at about the 2 hour mark from bending over the table where my scratch paper and textbook are.

[Vector]

So, question for you folks who do a whole bunch of math at a whack: How do you set up your work area? I start to get a sore neck and headache at about the 2 hour mark from bending over the table where my scratch paper and textbook are.

I often sit on the couch with a hard surface under my notebook, and the textbook next to me.  Other times, though, I end up sitting on the floor and working on the coffee table, usually with some sort of snack close by.

Generally, though, I fix that problem by massaging my neck, and usually I don't bend over very far: I keep my back straight and bend my neck a little, but other than that I try to keep my posture from eating me.

Other than that, I keep around blankets, water, various layers of clothing, a bunch of textbooks, a bunch of notebooks, scrap paper, and so on.  Typically, I sit with my back to the clock and don't play music.

[Heron TSG]

I have one of those 1.5 inch three ring binders for my math notebooks, and I use that as a writing surface while I migrate about my house, sometimes sitting on a couch, sometimes at a table, and sometimes just lying on the floor in random parts of my house. I like changing the scenery while I work.

[ed boy]

This is one thing that has rather annoyed me. The proof that the square root of two is irrational goes (informally) as follows:

let the square root of two be ^a/_b, where a and b are integers, and b is not zero, and a and b have no common factors.
(2)^0.5=^a/_b
2=^a2/_b²
a²=2b²
now since a² is a square number and is a multiple of two, a must therefore also be a multiple of two. we can then write a=2c, and a²=4c².
4c²=2b²
2c²=b²
now since b² is a square number and is a multiple of two, b must therefore also be a multiple of two.
but in the initial conditions, we chose a and b so that they had no common factors.
A paradox is created, so the square root of two therefore cannot be a rational number.

What is to stop one using this same method to "prove" that the square root of four, or nine, or sixteen is irrational?

[Vector]

Try writing a proof and see what happens.

[Vector]

In other words, because it's 3 AM and why not...

Spoiler: You wouldn't get a contradiction, is what happens (click to show/hide)

4^.5 = a/b, let a, b be relatively prime

4 = a^2/b^2

4b^2 = a^2

OK, replace a with 2c

4b^2 = 4c^2

b = c

So b = (1/2)a, but they have no common factors, i.e. 2b = a but 2b =/= 2ma for any m in N.

So b = 1 and a = 2 (application of the Fundamental Theorem of Arithmetic, basically)

?!

We didn't break mathematics?

[MaximumZero]

Did you just make 1 = 2?

Stop breaking my brain!

[Heron TSG]

No, B is 1, and A is 2, but they're not set equal to each other (I think.)

[ed boy]

4b^2 = a^2

OK, replace a with 2c

4b^2 = 4c^2

But doesn't the original proof rely on if N² is a multiple of x, N must also be?

[Christes]

Ahh, the road to becoming a professional mathematician... I can feel the paranoia riding strong again

Is it bad that I've taped my curtain to the wall so that no one can peak in through the gap in the side?

EDIT:

Spoiler: And for anyone who wants to share my pain: (click to show/hide)

Let S be a subset of T, and both equipped with a partial order relation.  Suppose that S is sup-dense in T.  Then prove that the canonical embedding is inf-continuous.

Spoiler (click to show/hide)

Oohh, oohh some advanced math.  Actually I'm not familiar with the terminology.  Let me guess: sup-dense means that each element of T is the sup of a subset of S.  Hmm, would inf-continuous be infimum-continuous?  Not quite sure what that would mean.  Probably preserves infima or something.  I'd think about it more, but I need to play with measurable functions today.

[Vector]

Yes, infimum-preserving under a morphism.  Good to see you, Christes

But doesn't the original proof rely on if N² is a multiple of x, N must also be?

Yes.  What's the problem?

[cowofdoom78963]

How do you divide?

[Heron TSG]

Well, let's say you have 5/5. Now, this looks easy, but you can make it simpler by instead multiplying by the inverse of the denominator, like so.

5/(1/5) - This is equivalent to the following, because any real integer (in this case 5) has an invisible denominator of one.

(5/1)(1/5) - Do some cross-multiplication by multiplying as such.

(25/5)x(1/5) - simplify this to get

(25/5)x(1/5) - Now, multiply the numerators and denominators.

(25x1)/(5x5) - simplify this to get

25/25 - simplify.

1/1 - this is equal to 1, because any number divided by itself is equal to 1.

Division problem solved without division.

^{If you were being serious, I both feel terrible and can help you actually do division if you want.}

[ed boy]

But doesn't the original proof rely on if N² is a multiple of x, N must also be?

Yes.  What's the problem?

In your explanation, you violate this, by doing this:

4b^2 = a^2

OK, replace a with 2c

4b^2 = 4c^2

This would be acceptable if, when you worked out your value of c it came to a multiple of 2. However, it does not, which means that a=2c cannot be a multiple of four.

[Vector]

a = 2c isn't a multiple of four.  a = 2.