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# September 24

## Injective and surjective polynomials

Let K be an algebraically closed field. Does a non-constant polynomial with coefficients in K necessarily define a surjective function from K to itself? Does a polynomial with coefficients in K defining an injective function from K to itself necessarily have degree 1? GeoffreyT2000 (talk, contribs) 22:58, 24 September 2016 (UTC)

Yes to both questions: (1) if f(x) is a polynomial and a is in K, then f(x)-a splits. (2) A degree n>1 polynomial that is one-to-one would have the property that f(x)-a is a perfect nth power for all a in K. In particular, f(x)=x^n, and x^n-1=(x-1)^n. This implies that n is a power of the characteristic of K, so f (x) is either the identity or is the Frobenius automorphism of a finite field. No finite field is algebraically closed, a contradiction. Sławomir Biały (talk) 23:44, 24 September 2016 (UTC)

# September 26

## Additive group homomorphisms of vector spaces which are not linear maps

Vector spaces are abelian groups under vector addition, and every linear map between vector spaces over the same field is also a group homomorphism of the vector spaces' additive groups. But what is an example of an additive group homomorphism between vector spaces over the same field that is not also a linear map? --Jasper Deng (talk) 08:27, 26 September 2016 (UTC)

Consider the complex numbers as a vector space over itself, and the map ${\displaystyle z\mapsto {\text{real}}(z)}$ from the complex numbers to itself.
You can also do things like a map from the reals to itself which is additive but not linear, but that's more complicated.--2406:E006:3D92:1:9C5A:A4ED:9EF0:6CC4 (talk) 10:09, 26 September 2016 (UTC)
An additive map between vector spaces over a prime field is linear, so you need more complicated fields. If you have a field extension L/K, then any non-identity automorphism in the Galois group Gal(L/K) is an example of an additive map from L to L that is K-linear, but not L-linear. (Complex conjugation is a classic example, similar to the one given by 2406 above). —Kusma (t·c) 12:37, 26 September 2016 (UTC)
The term linear is not really adequate in this context: one must be specific about the type of linearity. Every additive map is Z-linear. If you are referring to linearity with respect to a specific field in your question, the above answers cover it well. —Quondum 18:39, 26 September 2016 (UTC)
@Quondum: Forgot to clarify that the linearity is with respect to the field underlying both vector spaces (though Z, as a non-field, is out of scope of my question).
@Kusma: Right, I never thought to think of field extensions and their Galois groups. But how about a (nontrivial) example that is not linear with respect to scaling by members of any field? The example the IP gave, for example, is still real-linear.--Jasper Deng (talk) 02:59, 27 September 2016 (UTC)
An additive map is linear over the integers, and so it is (in characteristic p) linear over the prime field GF(p). If the characteristic is zero, then an additive map F can be easily shown to be linear over the rationals: ${\displaystyle F({\frac {n}{n}}x)=nF({\frac {1}{n}}x)}$ so ${\displaystyle F({\frac {m}{n}}x)={\frac {m}{n}}F(x)}$ for any x and any positive integers m, n. (The case of negative rationals is left to the reader). So the example you are looking for does not exist. —Kusma (t·c) 06:18, 27 September 2016 (UTC)
This really does go to show that being a vector space is quite a strong condition. Thanks. (For the case of negative rationals, again suppose m and n are positive integers, then Z-linearity allows us to conclude that F(-m/n x) = -m F(1/n x) using an argument similar to the above)--Jasper Deng (talk) 07:08, 27 September 2016 (UTC)
It seems a stretch to attribute this to bring a vector space, or to call it strong. Z-linearity of a function over a field already implies linearity with respect to the prime field. The definition of a vector space (requiring compatibility scalar multiplication) ensures that vector spaces inherit this property. Incidentally, Q-linearity (or Fp-linearity) for every field seems to be a weak condition to me: the combination of Z-linearity and inverses implies this for fields already. That no further linearity is implied by Z-linearity for any field makes it quite weak in my mind. Consider R as a vector space over Q: it is infinite-dimensional, so the number of Q-linear maps (on the field or vector spaces) that are not R-linear is huge. —Quondum 14:43, 27 September 2016 (UTC)

## A question on Geometry

In the given figure , D &E are the mid-points of sides AB and AC. If F & G are the points of tri-section of side BC .The Prove That :3∆FGH=∆ABC.See Figure At[ pic.twitter.com/suAt4FrSaD]

(I'm searching The method to prove .If That couldn't be given .Please provide the Hint) — Preceding unsigned comment added by Achyut Prashad Paudel (talkcontribs) 12:26, 26 September 2016 (UTC)

I'm not sure why you mention points D and E, since they don't appear in the statement of what is to be proven. Also, do you intend that H is the orthocenter, or something else? Loraof (talk) 13:18, 26 September 2016 (UTC)
• With H as the orthocenter, the assertion is true for several random examples I tried. The only way I can think of to prove it in general is not very elegant: Start by translating, rotating, and scaling the triangle to put its vertices at A(0, 0), B(1, 0), and C(p, q). It is easy to find the coordinates of F and G, but harder to find the coordinates of H. Then use the area formula in terms of vertices, given in Triangle#Using coordinates, to compute the two areas. Loraof (talk) 14:02, 26 September 2016 (UTC)
• Hint: how does the area of FGH relate to the area of BCH? (assuming H is the orthocenter)
My apologies--one of my examples I just worked out wrongly in too much haste, and the others were a special case (right triangles with hypotenuse BC and orthocenter H=A) for which the assertion is true. As Tigraan's hint shows, it is not true in general. Loraof (talk) 18:14, 26 September 2016 (UTC)

## At what times do the minute and hour hands of a clock make a straight line?

And noon and midnight don't count. That's when the minute hand and the hour hand are in the same location, and that happens ten other times too.— Vchimpanzee • talk • contributions • 18:16, 26 September 2016 (UTC)

See Clock angle problem#Equation for the angle between the hands. Loraof (talk) 18:22, 26 September 2016 (UTC)

Note that the question requires a "straight line", not superposition. Superposition provides a straight line, but there are other cases too (e.g. 6.00pm). You need to solve not just for an angle of 0 degrees, but also 180 degrees. — Preceding unsigned comment added by 79.71.175.99 (talk) 00:45, 27 September 2016 (UTC)

With a bit of thought, your can take the solutions for 0 degrees and transform them to get the answers for 180 degrees, without having to solve again. MChesterMC (talk) 08:52, 27 September 2016 (UTC)
Indeed. And the questioner should note that while it is important to understand and be able to set up and solve the equations, the times given in Clock angle problem#When are the hour and minute hands of a clock superimposed? can be quickly verified in your head. The minute and hour hands both move at a constant rotational speed, and thus move at a constant rotational speed with respect to one another. They are superimposed at noon, and then 11 more times over then next 12 hours, so this happens every 12/11 of an hour, or every 1 hour and 60/11 minutes, giving the those times of 0:00, 1:05.45, 2:10.90, etc. And from there, as MChesterMC suggested, it should be easy to see when the hands are opposite one another. -- ToE 12:31, 27 September 2016 (UTC)