• ## i am a stupid idiot

how do i learn how to math. I have a GED understanding of math in the very normal range.
Want to be a man, and not an ape.

• Try improving your understanding of logic first, if you haven't already. It is fundamental to all of mathematics.
Although I have not read it myself, I have heard good things about the "Book of Proof".

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• i have found some one willing to tutor me in algebra through calculus- where can I find textbooks (not pdf's, they require me to get a book) for cheap? Even if they're older, that would be fine with me.

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• Check out any thrift stores near you?

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• ## How to know if your product will sell X amount of people

LOL

• > sell X amount of people

This sounds like a slave trading product.

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• >>61
lol i mean sell to ORz

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• Ever since that other anonymous website made a major contribution to mathematics, I feel like we should be able to do it too. I vote that we work on the Riemann Hypothesis.

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• *with making a new structural theory, those are only good when they actually are more useful for solving some problems.

Places where anons could do a lot of good: chaos theory always seems to be in a bit of a shambles, there are huge open questions and a lot of it lacks rigor.

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• Please do! That would be awesome!
Not sure I can help...

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• >Ever since that other anonymous website ...

Which anonymous website?

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• ## ahhh

so you never really gonna be better consistently without any logic....

• x

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• ## Cantor's diagonal argument but for dimensional reduction

As I understand you can also apply Cantor's diagonal argument to a matrix (or well just write columns below eachother), thus proofing that you can always reduce an n-dimensional tensor to a 1 dimensional vector. Similarly, you can reduce an n-dimensional vectorspace to 1 dimension. For example, you could encode all the (natural) numbers in an n-dimensional vector in only one ("1D") number following Gödel, like this: 2^n_1*3^n2*...*p^n_i

• ## For all e>0, there exists a d>0 such that...

e-d statements are rad

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• $\eq$

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• some text $\LaTeX$ more text
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• percent signs are important

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• What's your favourite theorem?

1 reply omitted in this preview.

• Ummm, is it normal to know a lot of these theorems offhand?

• Either Gauss-Bonnet or the Theorema Egregium. Both are classical theorems related to Gaussian curvature, which is like a measure of how twisty a surface is at a point (sphere has positive curvature because it's bent in the same direction, hyperbola has negative curvature because it's bent in opposite directions, plane and cylander have 0 curvature because they're flat in a direction)

Gauss-Bonnet: if you start with a surface, and you stretch or bend it somehow, the integral of the curvature stays the same. This shows that, for instance, no matter how weirdly you stretch and bend a sphere, it stays a sphere (in some sense)

Theorema Egregium: if you keep distances between points fixed, then you also keep curvature fixed. This is why it is not possible to have a perfectly accurate map, because the paper is flat everywhere but the sphere is not flat anywhere. So any time you try to map any portion of a sphere, there will be some distortion of distances and area.

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• You guys are awesome. Unfortunately I don't know much. My brain is stained with dirty physics.

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• ## rather go to jail and die than work

can you use math to predict a product's sales? like maybe a certain design and format, and how will it get X amount of people? Maybe a certain mouse design like razor?

• ## Hello!

I love math!

• I love math too!

• Come to Mathchan (it's not ready yet).

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• ## Math books for kids

I've always wondered about how to get more kids into mathematics. I've even thought about writing math books aimed at children. To do that I want to read the existing literature out there. Let's compile kids' books and resources about teaching math

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• >>38

customer!

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• http://worrydream.com/SimulationAsAPracticalTool/

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• >>40
>Mathematics, as currently taught, consists of the manipulation of abstract symbols. For most people, the level of abstraction makes math unpleasant or unusable as a practical tool for exploring the problems of their lives.
>The simulations represent the problem concretely, without abstractions; provide a broader context, allowing a deeper understanding of the situation
This is interesting. The way I always saw it was that abstraction is the means by which we make logical thought, and was therefore the most important takeaway from mathematics.
Certainly learning practical applications is necessary as well, but being able to abstract those applications means they can be applied to more (possibly completely unrelated) situations...

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• ## Is this a thing?

1 reply omitted in this preview.

• Paywalled.
Can someone with access to the article, summarize shat the trick is ?

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• >>26
"This alternate method for solving quadratic equations uses the fact that parabolas are symmetrical.

For example, in this parabola:

y = x^2 – 4x – 5

The two solutions when y = 0 are the symmetrical points r and s, where the parabola crosses the x-axis.

The midpoint, or average, of r and s is the axis of symmetry of the parabola. We want r + s = –b, which happens when the average of r and s is –b ÷ 2. In this example: 4 ÷ 2 = 2.

The two solutions to the quadratic equation will be the axis of symmetry plus or minus an unknown amount, which we’ll call u. In this example:

r = 2 – u and s = 2 + u

To find u, we want the product of r and s to be equal to c, which in this example is –5. Rewriting r and s in terms of u:

r × s = –5

(2 – u) × (2 + u) = –5

Solving that yields 22 – u2 = –5 or u2 = 9, so u = 3 works.

The two solutions to this quadratic equation are 2 – u and 2 + u, or –1 and 5. In other words, this parabola intersects the x-axis when x = –1 and x = 5."

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• >>28
>Solving that yields 22 – u2 = –5 or u2 = 9, so u = 3 works.
Solving that yields 22 – u^2 = –5 or u^2 = 9, so u = 3 works.

Sorry. Copy-and-paste mistake.

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• ## The square root of 2 is irrational

Let $x$ be a positive number such that $x^2 = 2$. Assume for the sake of contradiction that $x$ is rational. Then there exist coprime non-negative integers $a,b$, with $b \ne 0$, such that $x = \frac{a}{b}$. Then we have $2 = x^2 = \frac{a^2}{b^2}$, so $a^2 = 2b^2$. Since $2$ is prime and $2$ divides $a^2$, $2$ must divide $a$. Say $a = 2c$ for some integer $c$. Then $4c^2 = a^2 = 2b^2$, so $2c^2 = b^2$, and therefore $2$ divides $b$. But then $a$ and $b$ are not coprime, which is a contradiction. Therefore $x$ is irrational. QED
• Thank you for your wonderful contribution! Always love to see some real maths here.

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