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December 23

1 + 1 = 0?

Long ago when I was in middle school I remember jotting down the following "proof" to my mild amusement:

(1)^{2}=(-1)^{2}

2\times \log(1)=2\times \log(-1)

\log(1)=\log(-1)

1=-1

1+1=0

2=0

Of course this can not be right, but can someone please point out the problem? 124.214.131.55 (talk) 01:21, 23 December 2008 (UTC)[reply]

Well, look at the lines, and see which is the first false one, and that ought to tell you where the error is. Is your first line true? Yes. What about the second? --Trovatore (talk) 01:26, 23 December 2008 (UTC)[reply]

I would assume that the premise (first line) is fine. The second line should logically be fine too as taking the log (or whatever) of both sides retains the equality. 124.214.131.55 (talk) 01:31, 23 December 2008 (UTC)[reply]

You're not taking Trovatore's advice. It's easier to determine whether a statement is true than whether reasoning is valid. Which line is the first false statement? Algebraist 01:40, 23 December 2008 (UTC)[reply]

Mathematical truth should be determined by valid, logical reasoning, not by preconceptions of what should or should not be. Thus I am hesitant on evaluating truth in the absence of logic. However, I will give it a try. Expanding the 1st line results in the identity 1 = 1. If I remember correctly, the second line becomes

0=(0+2\pi i)

. The real parts match, but the imaginary parts are not equal. Knowing that this is false still does not answer why. If the premise is valid, then doing an operation on an identity should maintain the identity. What is it about the logarithm that breaks this and why? 124.214.131.55 (talk) 02:32, 23 December 2008 (UTC)[reply]

Michael Hardy answers below, but we can get to the root of your trouble in a simpler way. OK, you agree that the second line is false. We certainly agree that logarithms of equal things must be equal. So what's left that could break? What manipulation have you done, in getting from line 1 to line 2, besides taking the logarithm of both sides? --Trovatore (talk) 02:51, 23 December 2008 (UTC)[reply]

See List of logarithmic identities#Powers for a version of

\scriptstyle \log(x^{y})=y\log x

that works when the involved logarithms have no real values, and that explains the extra

\scriptstyle 2\pi i

. -- Jao (talk) 16:36, 23 December 2008 (UTC)[reply]

Thanks Jao! That was exactly what I was looking for. I had been operating under the premise that

\scriptstyle \log(x^{y})=y\log x

, regardless of real vs. complex, which is apparently not correct. 124.214.131.55 (talk) 23:37, 26 December 2008 (UTC)[reply]

Taking the log of both sides preserves equality if there is a log to take. But your second line presumes there is a logarithm of −1. That's where it first gets problematic.

Now if we construe "log" to mean an extended version of the log function, for which "logarithms" of negative numbers are defined, then we hit another problem: Is it true that if log a = log b, then a = b? In other words, is this extended logarithm function a one-to-one function? Consider, for example, the fact you mentioned, that 1² = (−1)². Why not go straight from there to the conclusion that 1 = −1? The answer is that the squaring function is not one-to-one: two different numbers can both have the same square. It is true that two different numbers cannot have the same logarithm in the conventional sense of "logarithm", but that conventional sense says there is no logarithm of −1. For "logarithms" in this somewhat more unconventional sense, the logarithm function is not one-to-one and you have the same problem as with the squaring function. Michael Hardy (talk) 02:12, 23 December 2008 (UTC)[reply]

We have a pretty good article on invalid proofs if you want to see other examples. --MZMcBride (talk) 02:15, 23 December 2008 (UTC)[reply]

Latex with WinShell, regular expressions

I could not find any instructions in their table to replace \xy@ with \abc@ where @ is any non-letter. Be grateful if somebody can help. Thank you in advance. twma 08:41, 23 December 2008 (UTC)[reply]

Determinants of a Non-Square Matrix!!!

One day I was thinking about determinants. Determinants are defined as the magnitude of a matrix, and the magnitude of a vector(of course) is defined as the magnitude of a vector. Suddenly, I thought that (if determinant of matrix= magnitude of vector)if not only vectors of 1x1 dimension, that is the vector has the same number of rows and columns can have magnitudes then singular matrices can also have determinants!!! I worked out that you can get the determinant of a singular matrix like this. First see are there more columns or rows. treat the extra rows or columns as vectors and take the magnitudes of each of the vectors(columns if there are more columns and rows if there are more rows). Then the singular matrix will turn into a non-singular matrix and you can take the determinant of the square matrix!!! How can this be??? Is this possible???----~~The Successor of Physics~~ 14:43, 23 December 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

I suggest that you check your definitions of determinant and of singular matrix. It seems that what you mean by "Determinant of a Singular Matrix" has to be translated into: "norm of a non-square matrix", which is something definitely not worth an exclamation mark. I quote from the wikipedia article: "..determinant is also sometimes denoted by |A|. This notation can be ambiguous since it is also used for certain matrix norms and for the absolute value"!--PMajer (talk) 17:09, 23 December 2008 (UTC)[reply]

--

The determinant of a matrix is not its "absolute magnitude" if that term is taken to imply something that cannot be negative.

Sorry, I was mixed up! I corrected it.----~~The Successor of Physics~~ 10:49, 24 December 2008 (UTC)

And "non-singular matrix" is usually defined to be a square matrix whose determinant is zero. So non-singular matrices certainly do have determinants by conventional definitions.

Sorry, I was mixed up(again)! I corrected it(again).----~~The Successor of Physics~~ 10:49, 24 December 2008 (UTC)

You're really vague about how you're computing these determinants. You say "treat the extra rows as vectors", which of course is what they are, and take their magnitudes, which is no problem, and then you assert that somehow you get a determinant out of this. Specifically how you do that, you make no attempt to say. Michael Hardy (talk) 20:47, 23 December 2008 (UTC)[reply]

If you take away the extra rows/columns you get a square matrix, and there are bunches of methods saying how to get the determinants so I have no need to specify how you need to get a determinant out of this.----~~The Successor of Physics~~ 10:49, 24 December 2008 (UTC)

The other key point is that not only do you need to say how you calculate them but also how they are useful. Determinants are very useful numbers (they are invarient under lots of very useful transformations, they are multiplicative, etc.), what properties would your "determinant" have that would make it useful? --Tango (talk) 01:13, 24 December 2008 (UTC)[reply]

Superwj5, you seem to be saying that you can define a sort of determinant of a non-square matrix (this is different from "singular," by the way) by eliminating rows and columns from it so that you get a square one, and then taking the determinant of that. Is that right? Joeldl (talk) 11:14, 24 December 2008 (UTC)[reply]

Correct! That's what I meant!----~~The Successor of Physics~~ 12:47, 25 December 2008 (UTC) —Preceding unsigned comment added by Superwj5 (talk • contribs)

Then you will get many different possibilities, depending on which rows or columns you eliminate. The determinants that you get this way are called minors of your matrix. There is nothing particularly special, really, about choosing to eliminate the extra rows all the way at the bottom or the extra columns all the way to the right, rather than a different choice of rows or columns. The remaining rows/columns don't even need to be consecutive in order for this idea to make sense. (For example, a 5 × 7 matrix has 21 5 × 5 minors.) You can also get minors corresponding to smaller square submatrices if you eliminate both rows and columns. (For example, 3 × 5 matrix has 30 2 × 2 minors.) See Minor (linear algebra) for more on this. Joeldl (talk) 20:52, 25 December 2008 (UTC)[reply]

Angles of points on surface of sphere

Let us consider the points on the surface of a sphere centered at the origin. There are three axes, and each point has an angle of rotation about each axis (-pi to +pi). If we plot the points on the surface of the sphere in 3-d, with each axis being one of the angles of rotation, we get a surface bound by the cube centered at the origin with side 2*pi. This surface remains the same even if any two of the axes are interchanged. Could someone describe this surface, provide an equation, provide a plot, etc.? Thanks! --Masatran (talk) 16:10, 23 December 2008 (UTC)[reply]

Well, let's choose the angles α, β, γ to be the angular coordinate of the projections of the point onto the yz, zx and xy planes respectively. Note that this is not well-defined for the six "poles." If we first exclude any points where one of the coordinates is zero (and deal with those separately later), then we get z/y = tan α, etc. Your surface will be contained in the surface S with equation (tan α)(tan β)(tan γ) = 1. I don't think you get all of S. The octant x, y, z > 0 corresponds to that portion of S with 0 < α,β,γ < π/2. You'll have to play around with it to see where the other seven pieces are. Joeldl (talk) 17:07, 23 December 2008 (UTC)[reply]