Wikipedia:Reference desk/Archives/Mathematics/2008 October 22

= October 22 =

Perfect square
$$2^{182}+2^{1142}+2^{2100}$$ is a perfect square. Why? —Bkell (talk) 04:22, 22 October 2008 (UTC)


 * The idea is to rewrite it $$2^{182}+2 \cdot 2^{1141}+2^{2100}$$. Beginning to look familiar, right? -- Jao (talk) 05:40, 22 October 2008 (UTC)


 * Oh, I see. Thanks. —Bkell (talk) 06:54, 22 October 2008 (UTC)
 * Heh, a large part of my problem was that I had been writing 2100/2=1400. Oops... —Bkell (talk) 06:57, 22 October 2008 (UTC)
 * And what's $$3^{30}+3^{33}+3^{35}+3^{36}$$?

the answer is $$(3^{10}+3^{12})^{3}$$ --viju80 (talk) 07:37, 22 October 2008 (UTC) And by th way if $$2^a+2^b+2^c$$ is a perfect square, then it is obtained as expansion of some $$(2^d+2^e)^2$$ --PMajer (talk) 08:38, 22 October 2008 (UTC) And what about 49=1+16+32 you fool?--PMajer (talk) 15:55, 22 October 2008 (UTC)
 * I must say that after seeing all the hostility at the MOS talk pages, among others, it was very refreshing to see someone make a personal attack against himself for a change. -- Jao (talk) 18:47, 22 October 2008 (UTC)
 * Shows he's not a square :) Dmcq (talk) 20:01, 22 October 2008 (UTC)

Simultaneous linear equations
Im having trouble solving even some of the basic ones, here is the example given: Solve the following equation for x and y $$ax - y =c... [Equation 1]$$ $$x + by =d... [Equation 2]$$

Multiply Equation 1 by B $$abx - by = cb... [Equation 1']$$

Add Eq 1' and Eq 2 $$abc + x = cb + d$$ $$x(ab+1)= cb + d$$ $$x =\frac{cb + d}{ab + 1}$$

and therefore $$y=a\frac{d-c}{a-b}+c$$ $$=\frac{ab-bc}{a-b}$$

Here is the equation im trying (Is there a way to tell if i should use elimination or substitution, there are only two examples on the work and all the equations in y=ax+c y=bx+d form use substitution and the above form uses elimination?): ax+ y = c [Eq 1] x + by = d [Eq 2]

heres the best i got: ax+y = c [Eq 1] x +by = d [Eq 2]

Multiply ax+y by b abx+y = cb [Eq 1']

Add 1' and 2 abx + x = cb + d x(ab + 1) = cb +d $$x=\frac{cb+d}{ab+1}$$

The answer however is: $$x=\frac{d-bc}{1-ab}$$

cheers, Kingpomba (talk) 10:17, 22 October 2008 (UTC)


 * You made a mistake in multiplying [1] by 'b', and then in adding equations.
 * When multiplying 'ax+y = c' by 'b' you should get 'abx+by = bc'.
 * Then you should not add equations, but rather subtract them:
 * abx+by = bc
 * x +by = d
 * give:
 * (ab&minus;1) x + (b&minus;b) y = bc&minus;d
 * so:
 * x = (bc&minus;d) / (ab&minus;1)
 * HTH. --CiaPan (talk) 10:35, 22 October 2008 (UTC)


 * See also System of linear equations. --CiaPan (talk) 11:13, 23 October 2008 (UTC)

Is it true...
... that there are twice as many numbers as numbers?Mr.K. (talk) 12:13, 22 October 2008 (UTC)

2.20*2-2.20*4 —Preceding unsigned comment added by 122.50.244.47 (talk) 12:22, 22 October 2008 (UTC)

yes it is true. it is also true that there are half as many numbers as numbers.

it all depends on how you're comparing them. Here is a proof that there are as many positive integers as positive and negative integers combined: "count the positive and negative integers like so: 1) 1 2) -1 3) 2 4) -2 5) 3 6) -3 etc. By "counting" them you're establishing a 1:1 relationship with the positive integers.

You can also do the same thing for proving that there are as many numbers between 0 and 1 as there numbers greater than 0.

Establish two number lines like so: x  1 .  .  . 0 0...1...2..... You can always find one and only one line connecting number on the real number line with a number on the number line going from 0 to 1.... therefore there are just as many. (This is a geometric proof).

REMEMBER: when you say "there are as many" it means you can link them 1:1.

There's a famous "diagonal argument" showing there are as many rational numbers are integers. Just list all the rational numbers by their integer numerators and denominators in a grid: 1 2 3 4 5 6 7 8 9 .. 1 a b 2 c d 3 4 . . . So point a is 1/1, point b is 2/1, point c is 1/2, point d is 2/2. Obviously every positive rational, which by definition has an integer numerator and denominator, will be on this grid somewhere...

Then just count the number of items on this grid like so: 1 2 3 4 5 6 7 8 9 .. 1 a b f g o p 2 c e h n q 3 d i m r 4 j l s . k t . u. for clarity I've been counting with letters, but obviously you should count with the integers. Voila! Because every single point in the grid has a single integer, you've shown that there are as many positive integers as there are positive rationals! —Preceding unsigned comment added by 94.27.157.234 (talk) 13:07, 22 October 2008 (UTC)


 * Extra credit: can you do the same thing (or somethsing similar) as the grid above to establish that there are as many integers as there are real numbers? (Answer: No, because it's not true :) )  —Preceding unsigned comment added by 94.27.157.234 (talk) 13:14, 22 October 2008 (UTC)

Also, for more of this see cardinality and cardinal number. It's the same stuff as the discussion above, just worded in a standard mathematical way. The cardinality of the integers, for instance (which is the same as "the number of integers" as explained above), is denoted $$\aleph_0$$. The fact (again, proved above) that $$\aleph_0 = 2 \aleph_0$$ goes to show that the arithmetic rules we are used to for finite cardinals (like 5 or 31,248) don't work for infinite cardinals (like the "number of integers"). -- Jao (talk) 13:38, 22 October 2008 (UTC)

Parabola
Hi. A recent maths question I did prompted this query and my teacher couldn't help. Does the term 'parabola' refer specifically to $$y=f(x)$$ where $$f(x)$$ is a quadratic in x or does, for example, $$y=(x^2-4)^2$$ also describe a parabola as it is some function squared despite the fact that its highest power in x is four? Thanks 92.3.212.29 (talk) 16:49, 22 October 2008 (UTC)
 * Neither. The quartic you mention does not describe a parabola, and every non-trivial quadratic does, but there are other parabolas too. A parabola is a geometric object (a curve), rather than a function, and so any rotation of a parabola is a parabola, though most of them can't be described in the form y=f(x) at all. Algebraist 17:03, 22 October 2008 (UTC)
 * I can see how the curve $$y=x^4$$ might resemble a parabola (although, as Algebraist says, it really isn't one), but I don't see how you can mistake $$y=(x^2-4)^2$$ for a parabola, as it has three turning points. Gandalf61 (talk) 08:32, 23 October 2008 (UTC)
 * Well, I would grant the extenuating circumstances to whoever calls "parabola" the quartic $$y=(x^2-4)^2$$. If you think, the term "quartic", for a curve like this one, in the origin came just an adjective attached to "parabola". In fact, while the most common use of "parabola", and the most ancient too, is the one reported by Algebraist and Gandalf61 (i.e. the critical conic section), still it is true that the use of "parabola" in a generalized sense is also quite old. After Cartesius the standard name for many algebraic curves was "parabola" followed by adjectives, like "conica" or "quadratica" (for the classical one), "cubica", "quartica", "quadratoquadratica", "semicubica" ($$y^2=x^3$$), "quintica", "cubocubica"... And also many others adjectives, appealing to more geometrical aspects, like "nodata", "campaniformis", "cuspidata", "punctata", and referring to a classification that is quite obscure to me. However, many of these terms where still in use at the beginning of 1900, and sometime are also today (you know, just for showing off ;) --PMajer (talk) 14:48, 23 October 2008 (UTC)