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May 12[edit]

Why why why do I always have so much difficulty with analysis proofs?[edit]

Is the metric space $\left(\left\{f\in C[0,1]:\int _{0}^{1}|f(x)|dx\leq 1\right\},\int _{0}^{1}|f(x)-g(x)|dx\right)$ totally bounded! Widener (talk) 06:37, 12 May 2012 (UTC)[reply]

No it isn't. There's a very short proof using the fact that the metric space completion of the metric space is the closed unit ball in the Banach space space L¹([0,1]). Since no closed ball in an infinite dimensional Banach space is compact, this cannot be totally bounded. Basically, emulate the proof of this last fact (look in a book on functional analysis). Sławomir Biały (talk) 11:16, 12 May 2012 (UTC)[reply]

prime number[edit]

why is there an infinite number of prime numbers? 203.112.82.1 (talk) 20:38, 12 May 2012 (UTC)[reply]

Why? That's a deep philosophical question. Why is maths consistent? (or is it, Gödel?) It can be proved that there are an infinite number of prime numbers in the following way. First, assume that there is only a finite number of primes, which are A, B, C, D, etc. Now find the product of these numbers, and add 1, to get (ABCDEF... + 1) This number is not divisible by any known prime number, because it would leave remainder 1, and therefore must either be a new prime number, or be divisible by a new prime number (i.e. all numbers are all prime or the product of primes). Of course, now the list of known primes has been extended by one, this process (find product, add 1) can be repeated ad infinitum, each adding a new prime number to the list. As X (number of known primes) + 1 + 1 + 1 + 1 ... = ∞, there are an infinite number of primes.--Gilderien Chat|List of good deeds 20:49, 12 May 2012 (UTC)[reply]

You might want to take a look at the article Euclid's theorem which lists more proofs of this theorem. -- Toshio Yamaguchi (tlk−ctb) 22:15, 12 May 2012 (UTC)[reply]

Whats special about prime numbers? 203.112.82.129 (talk) 23:13, 12 May 2012 (UTC)[reply]

The fact that they are prime. Looie496 (talk) 23:18, 12 May 2012 (UTC)[reply]

I mean whats the use of prime numbers? i can call numbers ending in 8 cool numbers, but that doesnt make them special. 203.112.82.129 (talk) 23:20, 12 May 2012 (UTC)[reply]

As for applications, one is to communicate with aliens. Since they could find the prime numbers just as we can, and nature seems unlikely to generate primes, a radio signal with a prime number sequence ought to mean there is intelligence at the other end. StuRat (talk) 23:22, 12 May 2012 (UTC)[reply]

Any non-zero integer can be broken down into a product of prime numbers, and this decomposition is unique. For example 60 = 2×2×3×5. In the context of multiplication, primes are like the fundamental building blocks of all other integers. This gives them a lot of nice properties, for example the integers modulo n form a field only when n is prime. Number theory is basically an entire branch of math dedicated to the study of prime numbers. If you are interested in real world applications of primes, number theory has a lot of computer science applications. I will link to Number_theory#Applications but unfortunately it doesn't say much there. Rckrone (talk) 00:50, 13 May 2012 (UTC)[reply]

At a practical level, the most widely used methods for encrypting messages depend on prime numbers -- see integer factorization. Looie496 (talk) 02:41, 13 May 2012 (UTC)[reply]

They sometimes pop up unexpectedly, like in:

\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}=\prod _{p{\text{ prime}}}{\frac {1}{1-p^{-s}}}

(see Riemann zeta function) Ssscienccce (talk) 12:59, 14 May 2012 (UTC)[reply]

Some algorithms require two or more numbers whose greatest common divisor is 1; this is equivalent to saying they have no prime factors in common. —Tamfang (talk) 08:10, 15 May 2012 (UTC)[reply]

The pure mathematical reason for being interested in primes is that they show up in lots of interesting places, and there are some intriguing and important unsolved problems concerning them (for example, Goldbach's conjecture and the Riemann hypothesis). 81.98.43.107 (talk) 18:27, 16 May 2012 (UTC)[reply]

Simple groups[edit]

Let G a non-trivial finite group, and N one of the largest proper normal subgroups. My question: Is G/N always simple? --84.61.181.19 (talk) 22:05, 12 May 2012 (UTC)[reply]

Yes, as by the Third Isomorphism Theorem, normal subgroups of G/N are in canonical bijection with normal subgroups of G containing N. So N is a maximal normal subgroup of G (contained in no other proper normal subgroup) if and only if G/N is simple. You may also want to see this. --Sam^Talk 22:36, 12 May 2012 (UTC)[reply]

Not about simple groups: Let G a finite group and p a prime, such that o(g)=p for all g in G\1. My question: Is G always abelian? --84.61.181.19 (talk) 12:46, 15 May 2012 (UTC)[reply]

For p≥3, there are groups of order p³ which are not abelian, but have no elements of order p². See, for example,

Burnside, William (1897), Theory of groups of finite order, Cambridge University Press, p. 82

For p=2, the group must be Abelian, by simple algebraic manipulation. — Arthur Rubin (talk) 19:54, 16 May 2012 (UTC)[reply]

Slope of the Axes[edit]

We all know that two lines are perpendicular if the product of their slopes is -1. And we know that the slope of the x-axis is 0 and the slope of the y-axis is undefined. But the two axes are obviously perpendicular. So how can 0 × undefined = -1? Inter change able 22:33, 12 May 2012 (UTC)[reply]

If

m_₁m_₂ = -1

then:

m_₁ = -1/m_₂

If we plug in:

m_₂ = 0

then m_₁ is undefined, right ? StuRat (talk) 23:00, 12 May 2012 (UTC)[reply]

That's exactly the answer I got when I posed the question to a math-inclined person I know, but some others claimed it was wrong later on. Inter change able 23:07, 12 May 2012 (UTC)[reply]

What exactly did they claim was wrong with that logic ? StuRat (talk) 23:16, 12 May 2012 (UTC)[reply]

Answer 1: That's similar to asking, how can

{\frac {x}{x}}

be always equal to 1 if 0/0 is indeterminate. Mathematical laws have a range of applicability;

{\frac {x}{x}}

is equal to 1 whenever x is not 0, and the product of slopes of perpendicular lines is -1 whenever they are not horizontal and vertical.

Answer 2: More accurate then saying that

m_{1}m_{2}=-1

is to say that

m_{1}={\frac {-1}{m_{2}}}

. The slope of the y axis is undefined in the context of real numbers but more generally, it is an unsigned infinity. It is correct that

0={\frac {-1}{\infty }}

and that

\infty ={\frac {-1}{0}}

, though

0\cdot \infty

is indeterminate. -- Meni Rosenfeld (talk) 08:35, 13 May 2012 (UTC)[reply]

Here's an additional response. You wrote that "two lines are perpendicular if the product of their slopes is -1." That is correct but does not imply that lines are parallel only if the product of their slopes is -1. Perhaps the following argument, analogous to yours, will help you see how you'd gotten yourself confused: "Numbers are real if they are rational, and we know that

\pi

cannot be expressed as the ratio of two integers. So how can

\pi \neq p/q

be real?"—PaulTanenbaum (talk) 13:56, 14 May 2012 (UTC)[reply]

In the paragraph immediately above this one, "parallel" should be "perpendicular".→86.130.201.242 (talk) 18:52, 14 May 2012 (UTC)[reply]