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February 5

abs(X) < 1

I thought absolute value of X less than 1 is equivalent to -1 < X < 1 but it's wrong. What does abs(X) < 1 really mean?

I also often confuse the relationship where a square root of a value (say 9) is equal to +/- 3.

Is there any website that lists these tricky, confusing math relationships? —Preceding unsigned comment added by 142.58.43.102 (talk) 01:08, 5 February 2010 (UTC)[reply]

When X is real, the statements "

|X|<1

" and "

-1<X<1

" are equivalent. Why do you think they are not? Eric. 131.215.159.171 (talk) 02:21, 5 February 2010 (UTC)[reply]

[edit conflict] Among the real numbers,

|x|<1

is the same as

-1<x<1

. For the complex numbers, for example, instead of a line (from -1 to 1) you get a disk surrounding 0 when you write

|z|<1

. The square root function is, by convention, defined to yield the positive root, so that

{\sqrt {9}}=3,{\sqrt {9}}\neq -3

. Nonetheless,

(-3)^{2}=9

; when we are given that

x^{2}=9

, we can conclude (only) that

x\in \{-3,3\}

, which can also be written as

x=3\lor x=-3

or as

x=\pm 3

(the latter being a slight abuse of notation). As far as websites go, I of course suggest our articles on the absolute value and the square root; I'm not sure there's a centralized "confusing math" article (here or elsewhere), although there are pages like the Dr. Math FAQ. --Tardis (talk) 02:39, 5 February 2010 (UTC)[reply]

[edit conflict] The statement $|x|\leq 1$ means that the distance from x to 0 is less than one. However, since the distance from x to 0 is not dependent on the direction of x relative to zero (let us assume that the "direction" is "right" if x is positive, and "left" if x is negative), x could be any number strictly between −1 and 1. Also, if $|y|=x$ , $y=x$ or $y=-x$ ; thus intuitively, y is either x or the "reflection of x" about the origin/zero (think in terms of the number line if it helps).

With regards to the "square root of x" (where x is any real number), consider the "intuitive square root"; that is, consider ${\sqrt {x}}$ to be the positive square root of x. We know, of course, that $-{\sqrt {x}}$ is another square root of x. Why? Because if we square $-{\sqrt {x}}$ we get x, since $(-{\sqrt {x}})^{2}=(-{\sqrt {x}})(-{\sqrt {x}})=({\sqrt {x}})({\sqrt {x}})=x$ . Note furthur that both square roots of x satisfy the same relationship, and any number that satisfies this relationship is a square root of x. What is this relationship? Simply, the relationship is $|y|={\sqrt {x}}$ (by the argument in the paragraph directly above, this statement implies that $y={\sqrt {x}}$ or $y=-{\sqrt {x}}$ ).

If it was not obvious in my post, it is important not to memorize these relationships. Instead, attempt to understand their meanings geometrically; once this is done, their algebraic meanings will make sense to you as well (many people find it simpler to visualize algebra geometrically, and this is fine; everyone will have some preference even if they are competent at both). Disclaimer: I have used very "vague mathematical terms" rather than formalizm so that it is easy for the OP to interpret my post (so please do not criticize my post for its lack of formalizm!). PS T 02:51, 5 February 2010 (UTC)[reply]

How to know if a sequence is an arithmetic progression given the formula for its nth term

How does one know if a sequence is an arithmetic progression given the formula for the nth term? For example, the sequence for which the nth term is $5n+2$ is an arithmetic progression, but how do you tell from the formula that it's an arithmetic progression? Is there a formal proof that a sequence with a given formula for the nth term is an arithmetic progression?--220.253.218.157 (talk) 11:58, 5 February 2010 (UTC)[reply]

Arithmetic progressions are given by precisely those formulae that can be simplified to

an+b

, where neither

a

nor

b

contains

n

.

a

is then the difference between successive terms (constant by the def. of arithmetic progression) and

b

is the 0^th term. The tough part is finding a good simplification algorithm, unfortunately it can be difficult sometimes. — Pt _(T) 13:18, 5 February 2010 (UTC)[reply]

Another useful criterion is that a sequence a₀, a₁, a₂, ... is an arithmetical progression if and only if the differences of neighbouring terms a_n+1 − a_n are constant. — Emil J. 14:57, 5 February 2010 (UTC)[reply]

Marks out of ten

Frequently the homework etc of schoolchildren is assessed by giving it marks out of ten. I spoke to one parent who was concerned that his childs marks were lower when she moved to a different class in a different school. He did not understand that such marking is done in an intuitive manner without any marking schemes, so is going to vary from teacher to teacher, and is not a precise measurement. Is there any proof or evidence that using z-scores (ie adjusting the teacher's set of marks so that the mean and standard deviation were uniform) would produce more accurate results than intuitive marking? Thanks 78.144.242.151 (talk) 12:35, 5 February 2010 (UTC)[reply]

It depends. If the students in the two classes are more or less the same but the teachers' grading is different, then z-scores will give more accurate results. If the teachers are similar but the students are different, it will be less accurate. In general, to get meaningful results you will need to use equating tests which are given to samples of the students from all classes and graded independently of the class. -- Meni Rosenfeld (talk) 14:23, 5 February 2010 (UTC)[reply]

In practice, not only do the marks vary from teacher to teacher, the grade cutoffs do as well. Even in situations where the cutoffs are hypothetically fixed, teachers can give extra credit or use creative ways of finding the final "average". So it's usually not helpful to compare grades rfom different teachers on non-standardized examinations.

However, another source of lower marks in a new school may be that the student has not yet learned exactly what sort of things the new teacher is expecting. In that case, adjusting the z scores will not help at all – the student just needs to acclimate to the new environment. — Carl (CBM · talk) 16:38, 5 February 2010 (UTC)[reply]

What about the common situation where the same group of schoolchildren is taught and assessed by several different teachers. What would be best then? 78.146.215.222 (talk) 23:48, 5 February 2010 (UTC)[reply]

Yes, in this situation, standardisation of results gives a more accurate comparison (as mentioned by Meni above), and it is used in some schools (though disliked by teachers who don't understand it). Dbfirs 02:32, 6 February 2010 (UTC)[reply]

Is there any proof or evidence available anywhere that standardisation would produce more accurate results than intuitive marking? I have read the standardisation article. 78.146.77.179 (talk) 21:00, 6 February 2010 (UTC)[reply]

I didn't realise that you were asking about accuracy of marking. (I can't see that it would make much difference, except that marks out of ten are comparing pupils with perfection, whereas z scores are comparing with average). I was considering only comparison of marks between subjects. In the UK, examination boards always standardise marks before combining results from separate papers into one final result. Dbfirs 23:39, 6 February 2010 (UTC)[reply]

I'm not doubting your word, but do you have a source for that please? I'd like to investigate further. 78.146.77.179 (talk) 01:12, 7 February 2010 (UTC)[reply]

I'll try to find something on one of their websites. Their methods are not secret, but they don't publicise them widely. When I wrote "standardise", I didn't mean that they actually re-write results as z-values, just as adjusted marks, but it amounts to the same thing. Dbfirs 11:21, 7 February 2010 (UTC)[reply]

... (later) I think I might be wrong for the "Uniform Mark Scheme" system now used by UK examination boards. It achieves a similar result by determining grade boundaries on each paper and using linear scaling between these boundaries.[1] Dbfirs 13:21, 7 February 2010 (UTC)[reply]

Undefined terms in propositional logic

W(x,y) means x has visited y. Y's universe of discourse is all websites, x's universe of discourse is all students. $\exists y\forall z(y\neq David)\land (W(David,z)\to W(y,z)))$

Where did the z come from? I can't figure it out, and neither can my math teacher. Thanks! Sebsile, an alternate account of Saebjorn 15:57, 5 February 2010 (UTC)[reply]

z is just a bound variable. Since it appears only as the second argument in W(-,-), it must be a variable ranging over all websites. Similarly, y here is a bound variable ranging over students, in contrast to its meaning in the semantic definition of W. Algebraist 16:07, 5 February 2010 (UTC)[reply]

Correction: z is a bound variable if the missing left bracket is put immediately before (y≠David). If it's put right at the beginning, or between the two quantifiers, the situation changes. Algebraist 16:10, 5 February 2010 (UTC)[reply]

Right; whatever source this came from simply omitted a pair of parentheses that the usual rules of precedence say needs to be included. — Carl (CBM · talk) 16:34, 5 February 2010 (UTC)[reply]

Count again: it omitted half a pair, which definitely needs to be included. Algebraist 16:45, 5 February 2010 (UTC)[reply]

It seems to be a poor choice of notation. Since there is more than one universe of discourse, I assume that it is supposed to be multisorted logic. However, usually in multisorted logic there is some explicit syntactic sugar attached either to variables or to quantifiers to mark which sort they are supposed to come from. — Emil J. 16:35, 5 February 2010 (UTC)[reply]

In the low-level courses I have seen, it is typical to consider a single universe consisting of "all things" until the idea of a structure has been presented. This works out OK when one is only worried about translating from English to formal sentences and vice versa. Given the form of the sentence above, I assumed it comes from this type of setting. — Carl (CBM · talk) 16:41, 5 February 2010 (UTC)[reply]

Yes, I would also expect a low-level course (or even a high-level course, usually) to only work with one universe, as you say. However, then all the variables would have the same universe encompassing websites, students, and whatever else they need in the example, which is inconsistent with what the OP wrote. — Emil J. 17:04, 5 February 2010 (UTC)[reply]

Maybe all the students also happen to be websites? Rckrone (talk) 18:06, 5 February 2010 (UTC)[reply]

It does seem to be this way :( Could someone explain this in detail, in English? Oh, the course is "Discrete Mathematics and its Applications" Saeb(talkjorn) 04:02, 6 February 2010 (UTC)[reply]

If you're asking for a translation into English of the logical statement, I would say it means "There exists somebody who is not David who has visited every Web site David has." In other words, "There exists a person y, who is not David, such that David having visited a Web site z implies that y has also visited z." The variable z is needed to represent "some Web site," so that we can compare the visits of David to Web sites with the visits of y to Web sites. —Bkell (talk) 16:45, 6 February 2010 (UTC)[reply]

But can y be a person, even while its universe of discourse is websites? Saeb(talkjorn) 18:55, 6 February 2010 (UTC)[reply]

You can sum it up by the textbook beeing careless about specifying which universe of discourse the variables belong to. But even if you do not it sort of works if you massage the implication arrow a bit. Ie. we need the quantified implication to be true iff whenever W(David,z) is true then W(y,z) is true. And at the same time allow us to ignore the undified cases og W(David, Peter). This we can do by extending W to be false whenever the second argument is a person. Taemyr (talk) 19:15, 6 February 2010 (UTC)[reply]

There are two different meanings for y here. On the one hand, y is the second argument of the propositional function

W(x,y)

, in which case it stands for a Web site:

W(x,y)

means "Person x has visited Web site y." But in the statement

\exists y\forall z(y\neq {\text{David}})\land (W({\text{David}},z)\to W(y,z))

the variable y is being used as the first argument of the propositional function

W(y,z)

, so it must represent a person in this context, not a Web site. Saying that the universe of discourse of y is all Web sites is part of the definition of

W(x,y)

, where y is in the second position (and it is implied that the universe of discourse of the first argument, x, is all people). So in

W(y,z)

, the universe of discourse of z is all Web sites, whereas the universe of discourse of y is all people. It's the position of the arguments to the propositional function W that is important, not their names. —Bkell (talk) 19:41, 6 February 2010 (UTC)[reply]

Speed and Time problem

Recently I was in a sort of maths test, and I had this question "Mary has a destination to go to. She knows that if she walks at 10mph, she will arrive at 1pm, and if she walks at 6mph, she will arrive at 11am. She always leaves at the same time. What speed must she walk at to arrive at twelve noon?" It was a multiple choice question. I eventually went for 8mph, because it instinctively felt right. I did, however, have a feeling that it was a trick. I cannot remember all the other answers (One was √60, and the others were between 7 and 8). I had no idea of how I would work out the answer to the question when I was in the exam, and still do not. I am curious to know if it was as simple as it seemed, and if not how it could be worked out. Thanks in advance Chaosandwalls (talk) 23:40, 5 February 2010 (UTC)[reply]

She'll arrive later if she walks faster? That seems unlikely. Assuming you meant that if she walks at 6mph she arrives at 1 and if she walks at 10mph she arrives at 11, she must walk at 7.5mph to arrive at 12:

the OP forgot to mention that she was walking backwards... --Ludwigs2 22:51, 6 February 2010 (UTC)[reply]

Suppose she leaves at h o'clock in the morning. Then the distance she has to travel is 10(11-h) and also 6(13-h). So we have 10(11-h)=6(13-h), so 4h=32 and h=8, while the distance is 10(11-8)=30. So to get there at 12 she has to travel 30 miles in four hours, so she must go at 7.5mph. Algebraist 23:45, 5 February 2010 (UTC)[reply]

Ah I get it now, thanks very much Chaosandwalls (talk) 23:50, 5 February 2010 (UTC)[reply]

Also, the harmonic mean of 6mph and 10mph gives the correct result. -- Meni Rosenfeld (talk) 17:11, 6 February 2010 (UTC)[reply]

She's a remarkable person if she can walk at 6 mph for any distance, never mind 10. AndrewWTaylor (talk) 22:39, 6 February 2010 (UTC)[reply]