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Author Topic: Can't Think of a Breaking Bad Pun For the Title: Let's Do Some Math!  (Read 17794 times)

ankhtahr

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Well, we didn't have LHpital's rule yet, so we're not allowed to use it. Actually we didn't even have differetiation yet, so we're not allowed to use that either...
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PthariensFlame

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Well, we didn't have LHpital's rule yet, so we're not allowed to use it. Actually we didn't even have differetiation yet, so we're not allowed to use that either...


Really?  I can't think of how to solve, e.g., 25a or 25d without l'Hpital's rule; then again, my favorite calculus is definitely not the limit calculus.  :)
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What PhtariensFlame said. I'd say screw it and use it anyway rather than not getting any points at all. (Or, if you still have time to do so, ask your tutor or a jury of your peers if you can use it.)
Maybe you missed the introduction of it when you were sick?
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pwhodges

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Maybe the exercise is deliberately intended to get you to find out about the rule yourself - it's one style of teaching...
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Nah, it says:

"Beachten Sie bitte, dass Sie (...) nur den bisher behandelten Stoff verwenden drfen!"

Which basically translates to "only use the material that's already been used in the course". Or something like that. I almost know some German!
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Maybe the exercise is deliberately intended to get you to find out about the rule yourself - it's one style of teaching...
Sounds like being pushed of a cliff to see if the students evolve wings or not.

(Thanks, T.Pratchett!)
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ankhtahr

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Nope, we found solutions for both which were relatively simple without it.

I still remember our solution for the d) involving simply showing that lim cos(x) is 1, so lim cos(x) -1 is 0, so lim sin(cos(x) -1) is 0.

And as far as I know the rule will be introduced in a few weeks.
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ankhtahr

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Turns out it's hard to get back into Linear Algebra after being forced to skip a week.

Exercise Sheet

I have to read up on a lot right now. I don't even really know what vector spaces are.

I have until Monday evening for this sheet.
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I don't even really know what vector spaces are.


A vector space is a module over a field, with some appropriate additional laws.  :P
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Some informal thoughts:

You don't need to know what vector spaces are for the first one, just what linear (in)dependency is. Nevertheless, a vector space is a set of numbers which fulfills some axioms. A subspace is a subset of that set of numbers which also fulfills the axioms (ie is a vector space in itself).

Side note: A vector space is spanned by some vectors. Which means if you add vectors together or scale them up, you can get the whole vector space. Remember how you were calculating points in R in school? You needed a*(1,0) + b*(0,1) to get to any point (a,b). The set {(1,0), (0,1)} was a basis for R. Incidentally, the set {(1,0), (0,1), (13,37)} would also span R, but you'd never need to use (13,37) to reach your points. The vectors {(1,0), (0,1)} were a bare minimum you needed, that's why a basis is also called a minimum spanning system. There are many spanning systems for a vector space. For example, you could use {(4,0),(0,2)}, because you can cale them down to {(1,0), (0,1)} and proceed from there as usual.

If you only used {(1,0)}, then you could get all vectors which were on the x-axis, but no other vectors. All the vectors (x,0) are in a subspace of R. Incidentally, if you were to omit the 0 (because it's kinda redundant, right), you'd have the number line (that is, R). Congratulations! You discovered that R is a subspace of R!

Now, to some tasks.

1a) There are multiple approaches I can think of.
Because c1*v1+c2*v2+c3*v3=0 (I have substituted lambda with c) has at least one non-trivial solution, the set {v1, v2, v3} is linearly dependent. That means that you can "take away" at least one vector from it and they will still span the same space. <v1, v2, v3> = <v1, v2> = <v2, v3> = <v1, v3>.

Another way would be to start with a trivial case <v1>=<v2>. Then <v1, v2> = <v1> = <v2>. It would then follow from the equation that <v1> = <v3> = <v1, v3>. You'd then have to show that you get to the same conclusion even if <v1>!=<v2> (this is more formal, but harder and probably pointless).

1b) Would start with showing both directions separately.
=>: let (v_1, ...v_n) be a linearly indep. system. It then follows that it spans some subspace of V\{0}, let's call this subspace U. (v_1...v_s) then spans some subspace of U, let's call it U1. Similarly, (v_(s+1)...v_n) spans some subspace U2. Because (v_1, ...v_n) is lin. indep., every non-empty subset of it is also lin. indep. (this might be a bit of pain in the ass to prove. When in doubt, use induction). That is, for every s, {v1...vs} is linearly independent and {v_(s+1)...v_n} is linearly independent. In particular, for every v in U1, it is not in U2, except if it is 0. (I am doing it a bit sloppily right now.)
Because of this, the only element in U1 intersect U2 is 0.
<=: let's have <v1,...vs> intersect <v_(s+1), ...vn> = 0 for all s. Then it particularly follows that <v1> intersect <v2...vn> = 0. v1 is thus not in <v2...vn>. {v1} union {v2...vn}={v1...vn} is then lin. indep.

3b) Let V have n spanning vectors in it's basis and U have k spanning vectors in it's basis, with k<=n (ie U is k-dimensional). Then there are (n-k) vectors which are linearly independent (since they are a subset of a the linearly independent set of the basis vectors of V). They span a subspace W. None of the elements of W are in U, except 0 (does this look familiar?), and vice-versa. Thus, V\U=W.
I will leave the special case U=V as an exercise to you, because I am not sure if a complement can be an empty set.
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Reading that (for nostalgia's sake, though it's already more set theory than I ever did), I had the thought that my maths prowess was ultimately limited by the fact that I picture mathematical relationships geometrically - I cannot abstract the ideas to the point that I can handle them without that crutch, and so I am also limited to ideas that can fit into a three- (or maybe four-) dimensional space in my head.

That may not make sense to you, but I assure you it did to me.
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I, on the other hand, really struggle with geometry. I'm pretty good at maths, but going to linear algebra lectures were often pointless, because they modelled something that's basically just elements consisting of ordered sets of numbers as points in space, which became confusing.

The same thing goes for linear programming. It's basically the use of the very powerful simplex algorithm to solve a class of programming problems. The thing is, there's an algebraic way of looking at how it works and a geometric way of looking at how it works, and the geometric one is always used in explanations. I absolutely hate it. I can always see how the maths can model the geometry, but working with the geometry is just adding a layer of complication on something that's as simple as just numbers.
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ankhtahr

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So we're sitting here, working on the exercise sheet.

We have a solution for the 2a), but to me it seems overly complex, and formally wrong. I can see what the creator of the solution wanted to ge to, but I don't think you're allowed to write it that way.

(click to show/hide)

I don't know how it can be done easier though. (this is not my solution)
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ankhtahr

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So there's another linear algebra sheet due Monday. But as I don't have time to go to university on monday I need to be done with it on Friday. Damn.

Here's the exercise sheet.

I'm not happy about it.
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Skewbrow

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I've been gone for a couple weeks - nurturing that freshman course to a conclusion. My admiration for the level of freshman courses at your university is reinforced.

For 1A: As a homomorphism of groups f respects addition. Your task is to show that f(qv)=q f(v) for all rational numbers q and all vectors v. I would do this in steps as follows:
1) first prove it by induction whenever q is a positive integer (an example like this has probably been done).
2) a group homomorphism takes the additive inverse (=negative) to the additive inverse, so we can conclude that the claim holds whenever q is any integer.
3) next we do it in the case q=1/n for some positive integer n. Here a trick is needed. Write v=(n/n)v=n*((1/n)v). By step 1 we have f(v)=f(n*((1/n)v))=n f((1/n)v). Because W is a vector space over Q, this implies (multiply both sides by 1/n) that (1/n)f(v)=f((1/n)v), which is what we want for this step.
4) The general case q=m/n then follows by combining steps 2 and 3. Leaving this to you.

1B) Hint: Consider the Q-linear mapping f defined by f(1)=1, f(sqrt(3))=-sqrt(3). Show that it is not linear over Q(sqrt(3)).

2A) This you can do simply by using the subspace criterion. If phi1 and phi2 are two linear mappings sending all the elements of the subspace U to 0, then you can easily deduce that the same thing happens with the linear combination a*phi1+b*phi2, a and b in k.
2B) Barring suitable examples/auxiliary results in the course notes this makes your hands a bit dirty. Start with a basis of U, extend that to a basis B of V. You have probably covered the bit that in the finite dimensional case you get a basis for V* by mapping one of the elements of B, say b, to 1, and the rest to 0. Call that mapping phib. You should be able to show that the elements phib, b ranging over B, such that b is not an element of U, form a basis of U0.
2C) Many ways to do this. One is to find two linearly independent elements, v1 and v2,of V that are orthogonal to (1,1,2). Both of those give rise to an element in U0 by the recipe phii: x -> the inner product of x and vi, i=1,2. Another way (perhaps the intended way?) is to redo 2B in this specific instance: find such a basis of V that (1,1,2) is one of the basis vectors. Then describe in detail the two mappings in V* that map (1,1,2) and one of those other basis vectors to zero, and the remaining one to 1.
« Last Edit: 19 Dec 2013, 01:47 by Skewbrow »
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ankhtahr

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You won't believe how grateful we are for you Skewbrow. Without you we would have been doomed. With you, we've been a bit less doomed. It still took us a long time, but otherwise we wouldn't have figured out such nice solutions at all. My whole exercise-sheet-group has worked on this, and we wouldn't have made it without you.

Thank you.
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Loki

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This is not under time pressure, as it was for an assignment which has already been turned in, but:

Let's say you integrate a function over the reals, for example f(x)=x*exp(-x). The way I learned it, if you integrate a product of functions, you can call one function u(x) and the other v'(x) and then do integration/derivation voodoo with them, which I believe is the inverse chain rule method.

Now, my question: How do I decide which of the two functions should be u and which should be v'? Is there a rule of thumb?
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snalin

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Try one assignment, and if that's too hard, try the other. That was always my way of figuring it out.

As the ordering of the factors are irrelevant, which one you choose for u and v' is up to you - the result must always be the same.
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Actually, thereis a rule of thumb, and a mnemonic for it. 

The rule is that you want u to be something that gets simpler with derivation, and v' should be something that doesn't get much more difficult under integration. 

So, for example, if there's a logarithm, it's automatically u, because it can't be integrated without this trick anyway. 

So the rule of thumb; 

Choice of u should be

L - logarithmic terms (derivatives are negative powers of x, much simpler)
I - inverse trig terms (simpler derivatives, rational expressions, some with roots)
A - algebraic terms (polynomials, actually - the powers all get knocked down by one each time this is done)
T - trig terms (rarely simpler, but at least you're just getting other trig terms)
E - exponentials (no simpler, but no more complex when integrated, either, which makes it a good choice for v')

LIATE's not too hard to remember - it's almost a word.  If you have any two of these, use the first one that appears in the list should be used as u, the other one as v' and you'll usually get something simpler.  For your example,

u = x --> du = dx
v' = e^(-x) --> v = -e^(-x)

so int[x*e^(-x) dx] = (x)(-e^(-x)) - int[-e^(-x) dx] = -xe^(-x) - e^(-x) + C

BTW, this technique of reversing the terms in the product rule to get something easier is usually called "Integration by Parts", at least here in the US. 

And, lastly, this;

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Alt-text: "Eventually, a UN is set up. And then a lone rebel runs along the line of flags at the front, runs back to his base, and gets a kajillion points."
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Okay, so last time I did linear algebra was a year ago, so my memory is rusty.

I have a set of some five matrices in Q^(2x2). I am supposed to check whether this set a spanning set of Q^(2x2).
What I would do is transform a matrix ((a,b),(c,d)) with a,b as the first row and c,d as the second row into a column vector (a,b,c,d)^t, then use those column vectors to create a 4x5 matrix, which I would Gauss to check whether it has a rank of 4.

Iff it has a rank of 4, then my set is a spanning set.

Is what I am doing correct?
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...I'm pretty sure that's right. I just finished a semester on linalg, and your method seems sound. I haven't heard of Gaussing a matrix, though - we probably used a different term. You use row operations to reduce it to a row echelon form, right?
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Yes. I was using a colloquial used here for performing a Gaussian elimination (aka reducing it to row echelon form).

Thanks for doublechecking!
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Okay, so I've hit upon a fairly nasty inhomogenous second-order DE, and I'm stuck. Any help?

We have y'' - y' - y = cosh(x), where y(0) = 0 and y'(0) = 0.
I can get the homogenous part to be y = Ae-x + Be2x, but then I get stuck - I can't think of a substitution for y to solve the second part.
y = c cosh(x) + d sinh(x) doesn't work, so the answer can't be in terms of sinh and cosh.
I've tried separating cosh(x) into 1/2(ex + e-x), and treating it as two separate functions, but again I can't find a solution.

Ideas?
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Didn't check, but you should be able to handle that ex/2 part of hyperbolic cosine with the usual ansatz Aex. On the other hand the e-x/2 part requires an ansatz of the form Axe-x because e-x is a solution of the homogeneous DE.

But wait just a minute! Recheck the solution to the homogeneous part. The solutions of r2-r-1=0 are r1=(1+sqrt(5))/2 and r2=(1-sqrt(5))/2. Or perhaps you meant that your DE is y''-y'-2y=cosh(x) ?
« Last Edit: 19 May 2014, 07:13 by Skewbrow »
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Don't worry - I've figured it out  :-)
You were right, it was y'' - y' - 2y, I mistyped it.

On another note, I'm only just realising how different the names of maths terms are between countries - I've never even heard of an ansatz, but we use the same method.
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In this context this word just means "starting point" or rather "starting method" in German. It comes from the verb "ansetzen", which itself comes from the two words "an" / "daran" = "onto" and "setzen" = "to set"/"to place" (in the context of the word Ansatz), so it refers to the way you initially place a tool on a work piece.
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English math terms have borrowed in quite a few places from German. Consider, for example, eigenvectors and eigenvalues.
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Indeed, it has. Actually I only relatively recently learned that many (or at least some) English texts use the word 'Ansatz' like here. That is, guess the form of the solution with some unknown coefficients, and then try and see if it works for some choices of the coefficients.
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Someone on another forum I visit is doing some maths...:

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English math terms have borrowed in quite a few places from German. Consider, for example, eigenvectors and eigenvalues.
Physics terms as well: gerade/ungerade, brehmstrallung (probably spelled that one wrong...), etc.
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Can someone explain how to handle exponents in simple algebra equations?

According to the answer key, 25/2-23/2=23/2

But why??
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You can rewrite 25/2 as 2*2*21/2 (each time you multiply by two that adds 1 to the exponent).

So, in this case, 25/2-23/2 = 2*23/2-23/2 = 23/2 * (2-1) = 23/2
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In other words, it's not the exponents that needed handling.  It needed to be broken down into components in order to be solved. 
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So, just wondering, anyone here got any familiarity with first-order-logic?
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I know I took two semesters of logic, up through Godel's theorems, but that was back in the late 80's.  As for what I actually remember... well, let's see the question before I commit! 
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In other words, it's not the exponents that needed handling.  It needed to be broken down into components in order to be solved.

Got it, that makes sense. It could also be written as 2x - x = x(2 - 1). I remember fucking nothing of how to do math. How to multiply fractions?? Simplify quadratics?? What all the letters in SohCahToa stand for?? Ugh.

Robert is starting at community college this fall and we were trying to remember/learn how to do things for the placement test. He took the test last week and missed the "start out in a for-credit class instead of remedial" score by 2 points, so we spent a while on trig functions, and then there was only 1 trig question when he retook it today. *shrug* We'll see. There's no benefit in taking a class he's not prepared for, but spending a semester in a class that won't count in the end - because of two points - would suck.
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Loki

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I know I took two semesters of logic, up through Godel's theorems, but that was back in the late 80's.  As for what I actually remember... well, let's see the question before I commit!
Oh, I have no question at the moment - I just constantly hear the claim that barely any university outside of mine (at least in Germany) does mathematical logic as mandatory part of the curriculum. Thought I'd check if that was true. In case you are interested, the contents of the course are, roughly in that order:

* intro to Boolean logic
* completeness theorem for BL
* Horn formulas
* sequence calculus in BL
* compactness theorem for BL
* intro to first-order logic (FO)
* (finite and non-finite) axiomatization of structures in FO
* model-checking games
* theories (a thing so obscure there seems to be no definition of it on Wikipedia)
* Ehrenfeucht-Fraisse games
* sequence calculus in FO
* completeness theorem in FO
* compactness theorem in FO
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Carl-E

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I know I took two semesters of logic, up through Godel's theorems, but that was back in the late 80's.  As for what I actually remember... well, let's see the question before I commit!
Oh, I have no question at the moment - I just constantly hear the claim that barely any university outside of mine (at least in Germany) does mathematical logic as mandatory part of the curriculum. Thought I'd check if that was true. In case you are interested, the contents of the course are, roughly in that order:

* intro to Boolean logic
* completeness theorem for BL
* Horn formulas
* sequence calculus in BL
* compactness theorem for BL
* intro to first-order logic (FO)
* (finite and non-finite) axiomatization of structures in FO
* model-checking games
* theories (a thing so obscure there seems to be no definition of it on Wikipedia)
* Ehrenfeucht-Fraisse games
* sequence calculus in FO
* completeness theorem in FO
* compactness theorem in FO


 :laugh: :laugh: :laugh: :laugh: :laugh: not a goddamn thing there sounds familiar  :laugh: :laugh: :laugh: :laugh: :roll:

OK, one or two of the first few topics. 

Seriously, Math in the states, even at the graduate level vs. math in Germany is like a little kid trying to arm wrestle with The Hulk. 
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Skewbrow

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Logic? Lost me there. It was very popular at ND in the late 80s (and presumable before and after that), but I never caught the bug, so there's a Gdel size hole in my education. Peasant's logic is good enough to get a PhD, so...

Mind you, topics like that are popular here, too. More than half the research done at our math department is on problems that originated from computer science. We do not even seek to cover all of what is known as Core Math in US grad schools. I'm not at all sure that it is a good thing.
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GarandMarine

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Well I tell y'all what. I might just be a dumb jarhead who ain't had much use for learnin these many years, especially that there "arithmetic" but even I can tell you that that's pretty stupid.
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Barmymoo

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She's explained it in a ridiculous way, but that's how I add. It's not stupid, it's true - it is easier to add 10 and 5 than 9 and 6.
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Carl-E

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Barmy's absolutely right, the common core is all about the way we think about math to actually learn it.  Well, the way people who are fairly comfortable with math think about it - I liked it when the reporter mentioned that she had to "just memorize it".  Memorization works to some extent, but memorization backed by some understanding is easier, faster, and lasts longer. 

As far as explaining it, I get the feeling the teacher was trying to explain it at a meta-level.  As part of a news report, I get the feeling this was aimed at adults who are getting confused (and frustrated) when they see all this Common Core stuff that they never had to do. 

The other (and much bigger) problem with the transition to the Common Core is the same one that killed New Math in the 60's - the teachers.  They're doing a much better job educating the teachers on what they need to do, but a bad implementation (locally) of a good idea is just going to be a mess. 

A quick example; subtraction in the common core is initially taught by "counting up" from the smaller to the larger number.  That's the way most of us actually learned it, and I know several adults who are embarrassed that they still use that trick.  But it works...

So I saw an example the other day in some course materials  I was reviewing that took the example of 57 - 22, and counted up. 

By 5's. 

Now yes, you can land exactly on the result that way, but dammit, it's easier to go by 10's, it always is, it's a base 10 system!  Then do the last bit, the extra 5.  This is exactly the sort of crap that really makes a mess of things - it's not the common core itself (the ideas), it's bad implementations by teachers who aren't thinking about the math they probably learned through memorization!  And to top it off, they're often not flexible enough to see it any other way.  The same materials insisted that visual representations products always be row x column, so that

......
......
......
......

is 4 x 6, but

....
....
....
....
....
....

has to be 6 x 4! 

As someone who has always looked at things sideways, I pointed out that they damn well better accept both answers for both arrays, you never know what a kid's going to call a row! 

It's become another slow motion train wreck.  Toss in the whackos claiming it's federal mind control at work, and you've got a system doomed from the outset. 

Hell, why change anything, ever?  "It were good enuf fer mah semi-literate grampa, it's good enuf fer me (and mah kids)!"

[crawls back into his hermit-hole, muttering and cursing the whole way]
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GarandMarine

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Because government education is a waste of tax dollars at it's finest, and why let the market deliver efficient systems when we can waste time and money?

She's explained it in a ridiculous way, but that's how I add. It's not stupid, it's true - it is easier to add 10 and 5 than 9 and 6.


Goodie for you? Meanwhile 6+9 is 15. Want to have a brief race on simple addition and subtraction? Why weaken children by teaching them poor methods that require more work?
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How does it take more work to apply concepts you've already learned, rather than approaching everything as brand new? 9+6=15, yes, but kids have already been introduced to base-10, so it make sense to have them "find" the 10 that they're familiar with, and solve from there. I still solve 9+x by thinking of it as (x-1)teen. I don't try to pull the fact from my head fully-formed. I know perfectly well that I am stupid slow at basic arithmetic sometimes, so I use all the shortcuts I can.
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Carl-E

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Fine, then homeschool while you're also trying to work two - three jobs to feed and house them. 

Look, the foundng fathers were pretty clear (especially Jefferson) - an educated citizenry is in a democracy's best interest.  They left it to the states initially (as with so many other things), but it's grown far from the community-based schoolhouses.  The industrial revolution saw to that - automata need to be trained, after all! 

There's little evidence that a market based educational system will deliver at all.  You get what you pay for, and the poor will be left behind (again).  An equal opportunity to education is absolutely necessary, and a foundation of our country's entire system of government. 


As for your second point, this actually strengthens their ability to learn facts faster, remember them better, and extrapolate to do more than their parent's ever could.  No, we're not going to get a pack of Art Benjamins, but at least they'll stop with the fucking "I hate math" crap, because they'll actually understand what they're being asked to do. 

And Baindhe, you're far from stupid slow if you know and use the shortcuts! 
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GarandMarine

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There's little evidence that a market based educational system will deliver at all. 

Which is why private schools of any kind are consistently the bottom tier in all forms of testing me... oh wait. The free market can absolutely deliver better, cheaper and more efficiently then government. Hell any one could do better then the DOE that's been consistently failing American students for... is it three generations now?

I'm fine with some of my taxes (if we absolutely must be taxed) going to educating other people's larva. However I will be damned if I don't argue for that money to be used as efficiently and effectively as possible. Between that and my mother's career in education, I think it explains why I'm so anti-union.
 
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Barmymoo

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Private fee-paying schools are vastly disproportionately populated by children of wealthy families, who are already at an advantage for achievement. Stick the same kids in a public school and they'd still do well. Stick them in a school where there aren't any poor kids whose lack of educational support at home, or distracting personal circumstances, or simply generations of poor academic achievement telling in their critical thinking skills are holding everyone's progress back, and they will do even better. That's not proof that the market works. It's proof that society is incredibly unequal.

I absolutely don't deny that the American public school system is failing. It's an abysmal system and only seems to be getting worse. Public schools in other countries don't seem to have the same issue though. I would like to invite one of our Scandinavian forumites to input their experience in this discussion. It seems to me that most of the issues of publicly-funded services in the USA stem from not getting enough taxation funding, not too much. Obviously someone who has a blanket opposition to the mere concept of taxation wouldn't agree with that though.
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There's this really handy "other thing" I'm going to write as a footnote to my abstract that I can probably explore these issues in. I think I'll call it my "dissertation."

Loki

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Guys? We have moved into whole another kind of math entirely.
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