Math Challenge - July 2019

[Pi-is-3]

@fresh_42 Is my Q15 proof wrong?

 

[fresh_42]

@fresh_42 Is my Q15 proof wrong?

Sorry, no, and, yes, it was correct. It just slipped under the many posts about the Brownian motion and I hadn't seen it.

Nice coordinate system by the way. Pythagoras would have done, too, without complex numbers. Not that others think our high school problems need complex numbers to solve. Pythagoras is also what you called distance formula.

 

[fresh_42]

Oh, I think I understand now, partly..somewhat..hopefully..here we go again

Spoiler: Problem 8 - take 2 - still incomplete

Put $\alpha := \sqrt{1-\sqrt{3}}$ and $\beta := \sqrt{1+\sqrt{3}}$.
By Eisenstein the polynomial is irreducible, thus a minimal polynomial of $\beta$, hence
$$[\mathbb Q(\beta),\mathbb Q] = 4$$
(a basis is given by $1,\beta,\beta ^2,\beta ^3$). Now to find $[\mathbb Q(\alpha,\beta), \mathbb Q(\beta)]$. Note that $\alpha$ is algebraic over $\mathbb Q\subset \mathbb Q(\beta)$ so by relevant hint in #68
$$[\mathbb Q(\alpha,\beta),\mathbb Q(\beta)] = [\mathbb Q(\beta)(\alpha),\mathbb Q(\beta)] = 4,$$
so the field extension $\mathbb Q(\alpha,\beta) /\mathbb Q$ is of degree $16$ by the tower lemma. Not sure how to proceed at the moment. Does one observe how the automorphisms behave on the roots?

How can this be of degree $4$, if $2-\alpha^2=\beta^2\,$?

 

[fresh_42]

Mhh, the hint in #68 is inaccurate, then. I need to check the basics.. again. These side comments are too esoteric.

No, it just means that the minimal polynomial of $\alpha$ over $\mathbb{Q}(\beta)$ is not the same as over $\mathbb{Q}$. You have a basis $\{\,1,\beta,\beta^2,\beta^3\,\}$ of $\mathbb{Q}(\beta)$ over $\mathbb{Q}$ and $\alpha^2 \in \mathbb{Q}(\beta)$. So the only question is, whether $\{\,1,\alpha\,\}$ is linear independent over $\mathbb{Q}(\beta)$, which is a simple linear equation.

 

[nuuskur]

I understand my mistake: I missed the $\alpha \in K$ requirement. We don't have $\alpha\in\mathbb Q(\beta)$, hence argument that followed is gibberish.

Spoiler: Problem 8

Put $\alpha := \sqrt{1-\sqrt{3}}$ and $\beta := \sqrt{1+\sqrt{3}}$. The polynomial $x^4-2x^2 -2$ is irreducible by Eisenstein, hence minimal polynomial of $\beta$, which yields $[\mathbb Q(\beta),\mathbb Q] = 4$ with a basis of $\{1,\beta,\beta ^2,\beta ^3\}$. Observe that $\alpha^2 = 2-\beta ^2\in\mathbb Q(\beta)$ so it suffices to have $\{1,\alpha\}$ to generate $\mathbb Q(\alpha,\beta) / \mathbb Q(\beta)$ and they are linearly independent over $\mathbb Q(\beta)$. Thus
$$[\mathbb Q(\alpha,\beta),\mathbb Q] = 8.$$
Per definition, the Galois group is the group of automorphisms on $\mathbb Q(\alpha,\beta)$ that pointwise fix $\mathbb Q$. If $\tau$ is such an automorphism and $\alpha$ is a root, then $\tau (\alpha)$ is also a root. An automorphism is therefore determined by how it acts on the roots. This has to be (isomorphic to) a subgroup of $S_4$, one of order $8$ to be exact.

One also notices that the polynomial is even so the roots come in pairs: $\alpha, -\alpha$ and $\beta,-\beta$.

 

[member 587159]

I understand my mistake: I missed the $\alpha \in K$ requirement. We don't have $\alpha\in\mathbb Q(\beta)$, hence argument that followed is gibberish.

Spoiler: Problem 8

Put $\alpha := \sqrt{1-\sqrt{3}}$ and $\beta := \sqrt{1+\sqrt{3}}$. The polynomial $x^4-2x^2 -2$ is irreducible by Eisenstein, hence minimal polynomial of $\beta$, which yields $[\mathbb Q(\beta),\mathbb Q] = 4$ with a basis of $\{1,\beta,\beta ^2,\beta ^3\}$. Observe that $\alpha^2 = 2-\beta ^2\in\mathbb Q(\beta)$ so it suffices to have $\{1,\alpha\}$ to generate $\mathbb Q(\alpha,\beta) / \mathbb Q(\beta)$ and they are linearly independent over $\mathbb Q(\beta)$. Thus
$$[\mathbb Q(\alpha,\beta),\mathbb Q] = 8.$$
Per definition, the Galois group is the group of automorphisms on $\mathbb Q(\alpha,\beta)$ that pointwise fix $\mathbb Q$. If $\tau$ is such an automorphism and $\alpha$ is a root, then $\tau (\alpha)$ is also a root. An automorphism is therefore determined by how it acts on the roots. This has to be (isomorphic to) a subgroup of $S_4$, one of order $8$ to be exact.

One also notices that the polynomial is even so the roots come in pairs: $\alpha, -\alpha$ and $\beta,-\beta$.

I'm really looking for the exact isomorphism structure of the Galois group. There are only 5 groups of order 8 (up to isomorphism). Which one is it?

 

[nuuskur]

I'm really looking for the exact isomorphism structure of the Galois group. There are only 5 groups of order 8 (up to isomorphism). Which one is it?

The resolvent cubic is reducible and there are two complex roots, so it must be the dihedral group.

 

[JosephFG]

For #14,

Consider the limiting cases. With P and Q 0 away from the vertices, we have 50% red and 50% blue. As they move away from the vertices, the yellow area increases, so that when P and Q are both 1 away (the side of the rectangle), the diagram is all yellow. Yellow takes away from both red and blue, and the diagram is all straight lines, so, the rate at which yellow replaces blue and at which yellow replaces red are both consistent and constant. Therefore, it does not matter how far down the sides the points P and Q are. In every case, the ratio between red and blue is 1:1.

 

[fresh_42]

The answer is correct, but I do not follow you on your central statement.

Yellow takes away from both red and blue, and the diagram is all straight lines, so, the rate at which yellow replaces blue and at which yellow replaces red are both consistent and constant.

Why couldn't we have e.g. a parabola, with equal values at the extremes, but a maximum in the middle? Straight lines alone isn't sufficient. Of course it is in the end, but the constant rates have to be shown, either by analytic methods (easier) or by geometric theorems.

 

[JosephFG]

I began in my mind by considering the triangles formed by P and Q and the opposite sides of each, and I noted that no matter the placement of P or Q, the area of the triangle remains constant. Now I warn that I am a mere beginner in approaching Calculus, but, I think I grasp a little of it anyway, intuitively. The formalization of what I think I see needs improvement.

Anyway, this -- watching the triangles -- was the most obvious example of a parameter that does not change at all, in other words whose graph has slope zero. Here is the intuitive leap. I reasoned as follows:

Calculus is the mathematics of change. And it does not matter whether the change is in time or in space -- in this case it is change along the spatial dimension of the line segment. The rate of change is the derivative of the change itself. So I reasoned that, since everything we are looking at is straight lines, including the vector describing the path of P or Q as it transitions along the segment, their derivatives cannot but be horizontal lines, i. e. a constant rate of change. Two quantities simultaneously moving from 0.5 to 0 at constant rates of change must necessarily remain equal at every point.

If you can flesh that out with actual formalism, go right ahead. At the present juncture I don't know how to do that.

 

[fresh_42]

I began in my mind by considering the triangles formed by P and Q and the opposite sides of each, and I noted that no matter the placement of P or Q, the area of the triangle remains constant. Now I warn that I am a mere beginner in approaching Calculus, but, I think I grasp a little of it anyway, intuitively. The formalization of what I think I see needs improvement.

Anyway, this -- watching the triangles -- was the most obvious example of a parameter that does not change at all, in other words whose graph has slope zero. Here is the intuitive leap. I reasoned as follows:

Calculus is the mathematics of change. And it does not matter whether the change is in time or in space -- in this case it is change along the spatial dimension of the line segment. The rate of change is the derivative of the change itself. So I reasoned that, since everything we are looking at is straight lines, including the vector describing the path of P or Q as it transitions along the segment, their derivatives cannot but be horizontal lines, i. e. a constant rate of change. Two quantities simultaneously moving from 0.5 to 0 at constant rates of change must necessarily remain equal at every point.

If you can flesh that out with actual formalism, go right ahead. At the present juncture I don't know how to do that.

Well, as already mentioned, your idea is correct. Here we simply can calculate the areas and do not need to consider the rate of change:

Let's first label the areas and call the area of the square $X$.

Since height and baseline of both triangles equal the side length of the square, their area is half of $X$:
$$
A+E+H = \frac{1}{2}X = A+F+G = B+C+D+F+G = B+C+D+H+E
$$
This means $\dfrac{A}{B+C+D}= \dfrac{A}{\frac{1}{2}X -F-G}=\dfrac{A}{A}=1$

 

[JosephFG]

Actually, your last bit seems unnecessary. If A + F + G = B + C + D + F + G then we can subtract (F + G) from both sides and get A = B + C + D, or red = blue. Thank you for your help!

 

[TGVF]

Equation can be rewritten as: $f(x) = x(1+c^2_1+c^2_2) +x^3 +x^5+x^7 -c^3_1-c^3_2=0$
Where $f$ is an increasing function of $x$ since its derivative $f'(x) = 1+c^2_1+c^2_2 +3x^2 +5x^4+7x^6 >1$ for any real $x$. Furthermore, $x \rightarrow -\infty \Rightarrow f(x) \rightarrow -\infty$ and $x \rightarrow+\infty \Rightarrow f(x) \rightarrow +\infty$ , hence there is one and only one real root for this equation. QED

 

[nuuskur]

Spoiler: Problem 6 - some ideas

The representatives of the elements in the quotient ring are of the the form
$$a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1},$$
where $a_j\in \mathbb F_q$, so more compactly, they can be written as n-tuples $(a_0,a_1,\ldots, a_{n-1})$

If $e^1,\ldots, e^n$ is a basis for $\mathbb F_q^n$ then some selections of the basis elements span cyclic subspaces, for instace the entire space itself is of course cyclic. Denote the basis elements as follows $e^j := (e^j_1,\ldots,e^j_n)$.

Suppose $C := \left\langle e^{j_1},\ldots, e^{j_k}\right\rangle,\ k\leq n$ is a cyclic subspace. Now, considering the basis elements as polynomials we can consider the ideal generated by them (if $R$ is a commutative ring, then $r_1R + \ldots + r_kR \subseteq R$ is an ideal) and map $C$ to that ideal.

The subset of cyclic subspaces is a sublattice, because it's closed w.r.t $\cap, +$. Is there an isomorphism (of lattices) between this sublattice and ideals of the quotient ring?

I'm not even sure if any of the above makes sense or if I'm heading the right way.

 

[member 587159]

Spoiler: Problem 6 - some ideas

The representatives of the elements in the quotient ring are of the the form
$$a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1},$$
where $a_j\in \mathbb F_q$, so more compactly, they can be written as n-tuples $(a_0,a_1,\ldots, a_{n-1})$

If $e^1,\ldots, e^n$ is a basis for $\mathbb F_q^n$ then some selections of the basis elements span cyclic subspaces, for instace the entire space itself is of course cyclic. Denote the basis elements as follows $e^j := (e^j_1,\ldots,e^j_n)$.

Suppose $C := \left\langle e^{j_1},\ldots, e^{j_k}\right\rangle,\ k\leq n$ is a cyclic subspace. Now, considering the basis elements as polynomials we can consider the ideal generated by them (if $R$ is a commutative ring, then $r_1R + \ldots + r_kR \subseteq R$ is an ideal) and map $C$ to that ideal.

The subset of cyclic subspaces is a sublattice, because it's closed w.r.t $\cap, +$. Is there an isomorphism (of lattices) between this sublattice and ideals of the quotient ring?

I'm not even sure if any of the above makes sense or if I'm heading the right way.

You are certainly on the right track!

You correctly deduced that a representant of an element in the quotient ring can be written as

$$a_0+a_1X+ \dots + a_{n-1}X^{n-1}$$

Doesn't this give you an obvious map between $\mathbb{F}_q^n$ and $\mathbb{F}_q[X]/(X^n-1)$? Try to play around with this map: is it bijective? What is the image of a cyclic subspace? No need to use a basis.

 

[nuuskur]

doh.

Spoiler: Problem 6

Of course, the obvious map is obviously surjective and it is injective, because $1,x,\ldots, x^{n-1}$ are linearly independent. So, as abelian groups, they are isomorphic. Let $T: \mathbb F_q^n \to \mathbb F_q[X] / (X^n-1) =:R$ be the obvious map and $C$ be a cyclic subspace. Explicitly
$$(a_0,\ldots, a_{n-1}) \mapsto a_0 + a_1x + \ldots + a_{n-1}x^{n-1}$$

Spoiler: unnecessarily complicated

We show $T(C)$ is an ideal. It is a subgroup under $+$, because $C$ is a subspace. Take $(a_0,\ldots, a_{n-1})\in T(C)$ and $(r_0,\ldots, r_{n-1})\in R$. Then
$$(r_0,\ldots,r_{n-1})(a_0,\ldots, a_{n-1}) = r_0a_0 + r_0a_1x + \ldots + r_0a_{n-1}x^{n-1} + \left ( r_1a_0x + r_1a_1x^2 + \ldots + r_1a_{n-1}x^n + r_2a_0x^2 + \ldots + r_{n-1}a_{n-1}x^{2n-2} \right ) =: r_0a_0 + r_0a_1x + \ldots + r_0a_{n-1}x^{n-1} + c_0 + c_1x + \ldots + c_{n-1}x^{n-1} = (r_0a_0 + c_0, r_0a_1 + c_1,\ldots, r_0a_{n-1} + c_{n-1})$$
Since $C$ is a subspace we have $r_0(a_0,\ldots, a_{n-1})\in C$. This is the most technical bit of the problem, I think: is $(c_0,\ldots, c_{n-1})\in C$?

Let's switch gears. Fix an ideal $I\subseteq R$. Try to show, instead, that the preimage of $I$ is a cyclic subspace. It is a subspace, because $I$ is an ideal. Take $(a_0,\ldots, a_{n-1}) \in T^{-1}(I)$. We show $(a_{n-1},a_0,\ldots, a_{n-2})\in T^{-1}(I)$. Using the fact that $I$ is an ideal, it suffices to find a polynomial $f(x)$ such that
$$f(x) (a_0 + a_1x + \ldots + a_{n-1}x^{n-1}) = a_{n-1} + a_0x + \ldots + a_{n-2}x^{n-1}$$
which then implies $(a_{n-1},a_0,\ldots, a_{n-2})\in T^{-1}(I)$, which is what we want. If we take $f(x) = x$, we get
$$(a_0 + a_1x + \ldots + a_{n-1}x^{n-1})x = a_0x + a_1x^2 + \ldots + a_{n-2}x^{n-1} + a_{n-1}x^n \\ = a_{n-1} + a_0x + a_1x^2 + \ldots + a_{n-2}x^{n-1} + \underbrace{a_{n-1}x^n - a_{n-1}}_{=0}$$
and we're done.

Now, there is a one to one correspondence between the cyclic subspaces and ideals of the quotient ring. The map $C\mapsto T(C)$ is an isomorphism of lattices. Indeed, if $C_1,C_2$ are cyclic subspaces then $C_1\cap C_2$ (meet) and $C_1+C_2$ (join) are cyclic. Also $T(C_1\cap C_2) = T(C_1)\cap T(C_2)$, because $T$ is injective and $T(C_1+C_2) = T(C_1) + T(C_2)$, because it's compatible with addition.

 

[Pi-is-3]

I began in my mind by considering the triangles formed by P and Q and the opposite sides of each, and I noted that no matter the placement of P or Q, the area of the triangle remains constant. Now I warn that I am a mere beginner in approaching Calculus, but, I think I grasp a little of it anyway, intuitively. The formalization of what I think I see needs improvement.

Anyway, this -- watching the triangles -- was the most obvious example of a parameter that does not change at all, in other words whose graph has slope zero. Here is the intuitive leap. I reasoned as follows:

Calculus is the mathematics of change. And it does not matter whether the change is in time or in space -- in this case it is change along the spatial dimension of the line segment. The rate of change is the derivative of the change itself. So I reasoned that, since everything we are looking at is straight lines, including the vector describing the path of P or Q as it transitions along the segment, their derivatives cannot but be horizontal lines, i. e. a constant rate of change. Two quantities simultaneously moving from 0.5 to 0 at constant rates of change must necessarily remain equal at every point.

If you can flesh that out with actual formalism, go right ahead. At the present juncture I don't know how to do that.

I would have never thought of this. But using this method properly would be a very tedious task. First you need to consider original coordinates of P and Q as (t,0) and (0,t), find the intersection points, then take determinants to find area (even if you do it by short cut criss cross method, you will need to do it 4 times, which is extremely tedious. I know as my first attempt at this question was this but with direct coordinates). Then that would be equal. Might as well directly take the coordinates as (0,0.8) and (0.8,0) then do something like shear mapping to reduce area calculation, but the idea is cool.

 

[TGVF]

My solution to 11.
View attachment 246120

By the way i wanted to know whether can we solve the same integral but indefinite.If we can please provide a solution.

My solution to 11.
View attachment 246120

By the way i wanted to know whether can we solve the same integral but indefinite.If we can please provide a solution.

I agree but for the constant C which is not needed since we have here the definite integral $\int_0^{\pi /2} \, dx = \pi /2$.
No idea for the same integral I but indefinite form, .

 

[member 587159]

I agree but for the constant C which is not needed since we have here the definite integral $\int_0^{\pi /2} \, dx = \pi /2$.
No idea for the same integral I but indefinite form, .

You don't need to find an explicit indefinite form, as I explained earlier. It probably looks very nasty, if it exists at all.

 

[QuantumQuest]

Equation can be rewritten as: ...

Your solution is correct but the question has already been solved by @nuuskur.

Also, it is important when you solve a question not to forget to write for what question is the solution you post :)

 

[nuuskur]

I think #86 is not sufficient. We still need the other direction: image of a cyclic subspace is an ideal or else there are potential well-definedness problems of the supposed isomorphism of lattices. Right now, we have
$$I \text{ ideal} \implies T^{-1}(I) \text{ cyclic}$$
We'd also need
$$C\text{ cyclic} \implies T(C) \text{ ideal}$$

Spoiler: Problem 6 - finishing touches

Put
$$T : \mathbb F_q^n \to \mathbb F_q[x] / (x^n-1) =:R,\ (a_0,\ldots, a_{n-1}) \mapsto a_0 + \sum _{k=1}^{n-1} a_kx^{k}$$
Let $C \subseteq \mathbb F_q^n$ be cyclic. We show $T(C)$ is an ideal. As $C$ is a subspace and $T$ is compatible with addition, $T(C)$ is a subgroup. Take $r_0 + \sum_{k=1}^{n-1}r_kx^k\in R$ and $a_0 + \sum_{k=1}^{n-1}\in T(C)$. We must show their product is in $T(C)$.
$$\begin{align*} &\left (r_0 + \sum_{k=1}^{n-1}r_kx^k\right ) \left (a_0 + \sum_{k=1}^{n-1}a_kx^k\right ) \\ =&r_0\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1}\right ) \\ +&r_1\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1} \right )x \\ +&r_2\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1}\right ) x^2 \\ &\vdots \\ +&r_{n-1} \left ( a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1} \right )x^{n-1}. \end{align*}$$
As $C$ is a subspace, it is closed w.r.t multiplying by $r_k$. We also saw in #86 that multiplying by $x^k$ shifts the coefficients, but $C$ is cyclic, thus closed w.r.t shifting. All of the additives are in $T(C)$, therefore $T(C)$ is an ideal.

 

[member 587159]

I think #86 is not sufficient. We still need the other direction: image of a cyclic subspace is an ideal or else there are potential well-definedness problems of the supposed isomorphism of lattices. Right now, we have
$$I \text{ ideal} \implies T^{-1}(I) \text{ cyclic}$$
We'd also need
$$C\text{ cyclic} \implies T(C) \text{ ideal}$$

Spoiler: Problem 6 - finishing touches

Put
$$T : \mathbb F_q^n \to \mathbb F_q[x] / (x^n-1) =:R,\ (a_0,\ldots, a_{n-1}) \mapsto a_0 + \sum _{k=1}^{n-1} a_kx^{k}$$
Let $C \subseteq \mathbb F_q^n$ be cyclic. We show $T(C)$ is an ideal. As $C$ is a subspace and $T$ is compatible with addition, $T(C)$ is a subgroup. Take $r_0 + \sum_{k=1}^{n-1}r_kx^k\in R$ and $a_0 + \sum_{k=1}^{n-1}\in T(C)$. We must show their product is in $T(C)$.
$$\begin{align*} &\left (r_0 + \sum_{k=1}^{n-1}r_kx^k\right ) \left (a_0 + \sum_{k=1}^{n-1}a_kx^k\right ) \\ =&r_0\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1}\right ) \\ +&r_1\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1} \right )x \\ +&r_2\left (a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1}\right ) x^2 \\ &\vdots \\ +&r_{n-1} \left ( a_0 + a_1x + a_2x^2 + \ldots + a_{n-1}x^{n-1} \right )x^{n-1}. \end{align*}$$
As $C$ is a subspace, it is closed w.r.t multiplying by $r_k$. We also saw in #86 that multiplying by $x^k$ shifts the coefficients, but $C$ is cyclic, thus closed w.r.t shifting. All of the additives are in $T(C)$, therefore $T(C)$ is an ideal.

Your solution seems fine to me now. Well done!

 

[Not anonymous]

Solution to problem 13.

Spoiler

Proof by contradiction. Suppose all the 3 products, $u(1 - v), v (1 - w) and w(1 - u)$ are greater than $1/4$. Then it follows that:

$u(1 - v) > 1/4 \Rightarrow u > 1/4(1-v) \Rightarrow (1 - u) < 1 - 1/4(1-v)\;\;\; \mathcal (Eq. 1)$

$v(1 - w) > 1/4 \Rightarrow v > 1/4(1-w) \Rightarrow (1- w) > 1/4v \Rightarrow w < 1 - 1/4v\;\;\; \mathcal (Eq. 2)$

From (Eq. 1) and (Eq. 2), it follows that
$w(1-u) < (1 - 1/4(1-v)) \cdot (1 - 1/4v) = ((3 - 4v) / 4(1 - v)) \cdot (4v - 1) / 4v \\ = (3 - 4v) (4v - 1) / 16v(1-v) = (12v - 3 - 16v^2 + 4v) / 16v(1-v) \\ = (16v - 16v^2 - 3) / 16v(1-v) = (16v(1-v) - 3) / 16v(1 - v) \\ = 1 - 3 / 16v(1-v)$

It it easy to see that for RHS to have maximum value, the denominator of the subtracted value should be maximal. It is trivial to show that $v(1-v)$ reaches maximum when $v\; =\; 0.5$ and therefore the maximum value of RHS would be $1 - 3 / (16 \times 0.5 \times (1 - 0.5)) = 1/4$. It follows that $w(1 - u) < 1/4$, disproving the initial assumption that all 3 products could be greater than $1/4$

 

[mfb]

You can save some steps in that proof:

Spoiler

Assume there are u,v,w such that all three expressions are larger than 1/4. Then the product of the three expressions must be larger than 1/4³ = 1/64.
That product is u(1-u)v(1-v)w(1-w). All factors are positive, to find its maximum we can maximize u(1-u), v(1-v) and w(1-w) individually. The maximum is 1/4 each, the product is 1/64, contradiction.

 

[nuuskur]

I gave my take on 13 in #42. Relies on how I like to prove statements of the form $A_1\lor A_2\lor \ldots \lor A_n$. Equivalently, one can prove
$$\neg A_1\land \ldots\land \neg A_{n-1} \implies A_n.$$

 

[fresh_42]

My proof uses $u(1-u)=\frac{1}{4}-(u-\frac{1}{2})^2< \frac{1}{4}$ since squares are positive.

 

[Pi-is-3]

My proof uses $u(1-u)=\frac{1}{4}-(u-\frac{1}{2})^2< \frac{1}{4}$ since squares are positive.

Wow. That is the simplest proof for that question!

 

[mfb]

My proof uses $u(1-u)=\frac{1}{4}-(u-\frac{1}{2})^2< \frac{1}{4}$ since squares are positive.

$\leq$ as u=1/2 is possible - squares can be zero. I didn't go into detail with that part as it is very easy to show and a well-known result, too.

 

[Pi-is-3]

$\leq$ as u=1/2 is possible - squares can be zero. I didn't go into detail with that part as it is very easy to show and a well-known result, too.

I think that's just a typo.

 

[fresh_42]

$\leq$ as u=1/2 is possible - squares can be zero. I didn't go into detail with that part as it is very easy to show and a well-known result, too.

Hey, I was lazy: "<" is on the keyboard $"\leq"$ is not. These details are trivial and left to the reader ;-)

 

[nuuskur]

Hey, I was lazy: "<" is on the keyboard $"\leq"$ is not. These details are trivial and left to the reader ;-)

That reminds me of a statistics course lecture. The lector said something like "..and since $f$ is injective, it is invertible..". Similarly, in algebra, saying something is a subgroup of $G$ is to be read as "is an isomorphic copy of a subgroup of $G$".

 

[fresh_42]

That reminds me of a statistics course lecture. The lector said something like "..and since $f$ is injective, it is invertible..". Similarly, in algebra, saying something is a subgroup of $G$ is to be read as "is an isomorphic copy of a subgroup of $G$".

or omitting "almost everywhere" in measure theory.

I have also seen isomorphic for monomorphic (here, not in books), which I really dislike.