7 + 5 = x, solve for x
is easy7 + 5 =
i would stare at for a couple of minutes then either insert the x myself and solve it or get out a calculator
Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?
But you do have r. The circle's area is 9 mi^{2}, so the radius is sqrt(9/pi).
I know, the 9th hour is not a pretty one...
hate/love :psyduck:
(http://latex.codecogs.com/png.download?%7B%5Ccolor%7Bwhite%7D0.5%5Cgeq%5Cfrac%7B%5Cleft%7Cx%5Cright%7C%7D%7B1+x%5E2%7D%7D)
I wish to use this opportunity to declare my hatred towards absolute-value inequalities.That said, multiply both sides by the denominator. This does not change the unequality sign, because 1+x² is >0 in real numbers. Then use this site:
Damnit, I just noticed that (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%20%5Cbinom%7Bn+1%7D%7Bk%7D%3D%5Cbinom%7Bn%7D%7Bk-1%7D+%5Cbinom%7Bn%7D%7Bk%7D%7D) is of course only valid for (http://latex.codecogs.com/png.latex?%5Cinline%20%5Clarge%20%5C%5C%20%7B%5Ccolor%7BWhite%7D%201%5Cleq%20k%5Cleq%20n%7D). So it does not help me, as in the sum there is k=0.
Apply the identity (http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn+1%7D%7Bk+1%7D%20%3D%20%5Cbinom%7Bn%7D%7Bk%7D%20+%20%5Cbinom%7Bn%7D%7Bk1%7D%7D) to all terms of the sum, except the first and the last:
(http://latex.codecogs.com/png.latex?%5Cinline%20%7B%5Ccolor%7BWhite%7D%20%5C%5C%20%5Cbinom%7Bn+1%7D%7B0%7D%20*%20%28-1%29%5E%7B0%7D%20+%20%28%5Cbinom%7Bn%7D%7B0%7D+%5Cbinom%7Bn%7D%7B1%7D%29*%28-1%29%5E%7B1%7D%20+%20...%20+%20%28%5Cbinom%7Bn%7D%7B0%7D+%5Cbinom%7Bn%7D%7B1%7D%29%20*%20%28-1%29%5E%7Bn%7D%20+%20%5Cbinom%7Bn+1%7D%7Bn+1%7D%20*%20%28-1%29%5E%7Bn+1%7D%7D)
Several years ago, A/Prof Barry Hughes set a ﬁrst-year exam question which asked for a proof that there is no largest prime number.The following was sent to him by one of the graduate students who helped with the marking:... That took a while to copy over, but I think you'll agree it was worth it.
Hi Barry, just for educational value, I compiled a list of the most incorrect answers to question B4(b). It is entitled "15 good reasons why Pure Mathematics is not taught to first year students."These are all verbatim answers from the 150-odd papers I marked, i.e. about one in ten. My favourite is deﬁnitely number 10; I think that is quite ingenious.
- Proof by example: “Let x be the largest prime. Then x = 91 but 91 + 6 = 97, which is prime. Therefore, 91 cannot be the largest prime number. Therefore there is no largest prime number.”
- Proof by oddness: “If n is the largest prime number, then n is odd. Then (n + 1)=2 is even. Therefore, (n + 1)=2 + n is odd. But (n + 1)=2 + n is not divisible by any number except itself. As it is bigger than n, the assumption is wrong, by contradiction.”
- Proof by intuition: “Prime numbers are integers that can be divided by themselves only; prime numbers are odd with the exception of 2. By intuition as n -> infinity, there will always be an odd number that cannot be divided by any other number besides itself.”
- Proof by sqrt(2): “that p is divisible by q, i.e. p/q = 2r, where r is an even number. Then p = 2qr so p^{2} = 4q^{2}r^{2}. But r^{2} does not exist and q! = 1. Therefore, q must exist. Since q exists, p must be divisible. Therefore, by contrary, there is no largest prime number.”
- Proof by superinduction: “2 is a prime number. Now assume N is the largest prime. But then N + 1 exists and is also prime. Therefore, by induction there is no highest prime number.
- Proof by the previous question: “Suppose N is the largest prime. Then let N = (n^{2})/2. Therefore n = sqrt(2N). But from above (1 + 2 + 3 + ... +n) > N. Hence, there is a larger prime number than N.”
- Proof by tutorial question: “Let m, n be two integers with m > n + 1. If k is even, m^{k} + n^{k} cannot be expressed in terms of (m + n)(polynomial in m and n) and so is prime. Therefore, as m and n can be any numbers, there is obviously no largest prime number.”
- Proof by having no idea what a prime is: "Say the largest prime possible is x, then 2x is also a prime since the statement is true for all natural numbers."
- Proof by experimental data: "Suppose n is the highest prime. Then 2n-1 is also prime. But 2n-1>n so there is no highest prime (Check: 2*2-1=3, 2*3-1=5, 2*5-1=11, 2*11-1=23, so true)"
- Proof by subscript: “If there is a highest prime, we can number all the primes p_{1}, p_{2}, ... , p_{n}. But as there is no highest natural number, there is always an n + 1 so there must be a p_{n+1}. Therefore, there is no highest prime.”
- Proof by inﬁnity: “Let n be the highest prime number. But infinity is greater than all numbers so infinity > n. If n is the highest prime this would mean infinity has factors. Therefore, we have a contradiction.”
- Proof by reverse logic: “All prime numbers are odd. Suppose there were a highest prime. Then we have a highest odd number. But if 2k+1 is the highest odd number, then 2k+3 = 2(k+1)+1 is also an odd number. Therefore, we have a contradiction and therefore we have a contradiction."
- Proof by denial: “Assume there is a largest prime M. We can add 1 to M until we get another prime number N (M + 1 + 1 + 1 + ... + 1 = N). But then N > M. Therefore, M is not the largest prime number, so there is no largest prime number.”
- Proof by formula: “As prime numbers are derived via the formula, we can assume it works for n = k giving the highest prime number. But then it also works for n = k + 1, so there is no highest prime number.”
- Proof by continuity: “Let x be the largest prime number. Then x > all other primes. But then (x + n), the next prime number, does not exist. However, numbers are continuous and so (x + n) does exist. Therefore, there is no x.”
— George Doukas
Prove: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.
QuoteProve: Each convergent series over K \in {\mathds{R}, \mathds{C}} is bounded.
Is there any sensible answer except "well. duh, if it converges to x, it is bounded by x"?
(Topology is almost nothing but a collection of counterexamples... I got million of 'em!)
:x
Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?
Is. "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics". Many words end in s without being plural.
Let G be a group and H1, H2 subgroups of G.
Show that H1 union H2 is a subgroup or find a counterexample
You are doing weird stuff.
Re 2b)QuoteLet G be a group and H1, H2 subgroups of G.
Show that H1 union H2 is a subgroup or find a counterexample
Don't really think anyone here can help me with it, if I don't give a translation. I don't have the time to give a translation right now.If the topic of the sheet is something like the first course in abstract algebra (judging from Loki's post), I am qualified. I don't fully trust my mathematical German, but I have read a number of German math books (though it was a while ago). The problem is that my browser is convinced that following up your link is a bad idea, risky even.
Maths is, right? I mean, maths is written in plural form, but it is still just a single thing, right? Or is it maths are sexy?
Is. "Maths" is not the plural of "math"; it's a different abbreviation of "mathematics". Many words end in s without being plural.
Depends. It's "Maths" in the UK, and "Math" on the other side of the pond (or should I say elsewhere?).
The problem is that my browser is convinced that following up your link is a bad idea, risky even.
Assignment 1: Let G and H be groups with respective neutral elements e_{ G } and e_{ H }
Let f be a homomorphism of groups from G to H. Show that:
c) Let p: G -> G/ ker( f ) be the canonical projection mapping the element g to its coset g ker(f) (sorry, I don't know
what "coset" is in German). Then there exists a unique homomorphism f from G/ ker( f ) to H such that
f = f o p (using underscore instead of the overline bar, sorry). Furthermore, f is injective and its image
is equal to that of f.
Assignment 4: Let M be a set and P(M) its power set. For all elements A, B of P(M) define the
symmetric difference (Can't draw the Delta symbol here. I'm using '+' instead, because this will become
the addition of the ring you are creating. Similarly 'n' denotes intersection when standing alone).
A + B := (A U B) \ (A n B).
Show that (P(M),+,n) is a ring. You are allowed the assume the associativity of the symmetric difference.
Assignment 1:
A) Show that the set Q(3^{1/2}) of numbers of the form a+b3^{1/2}, with a and b ranging over Q, is a field with respect to the usual addition and multiplication of real numbers.
B) Show that the mapping that sends a+b3^{1/2} to a-b3^{1/2} is a well-defined automorphism of fields.
Hint: You are allowed to assume without proof that 3^{1/2} is irrational, i.e. not an element of Q.
Prove: every polynomial function p: R -> R (with R as reals) of an odd degree has at least one x-intercept.
Well, we didn't have L’Hôpital'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...
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.
I don't even really know what vector spaces are.
(http://imgs.xkcd.com/comics/borders.png)
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.
In other words, it's not the exponents that needed handling. It needed to be broken down into components in order to be solved.
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:
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
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.
There's little evidence that a market based educational system will deliver at all.
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?
It's like teenagers at a catholic sleepover party: they can get as close to each other as they wish, as long as there is no touching.
Welp, my work here is done!o.O I am far from being the most math adept person on the forum (not counting you, of course). We have physicists amd actual rocket scientists here.
I figured if anyone would crank it out, it would have been you... and Fermat's sum of two squares theorem was a nice easter egg!
Have fun proving it...Welp. Uh. I can do one direction. That's already half of the work, right? :-D The rest is left as an exercise.