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October 12

Sum of angles of a triangle

The sum of the angles of a triangle add up to 180 degrees. Where does this come from? How is it proven that this is always the case. --Polyknot (talk) 03:21, 12 October 2017 (UTC)[reply]

I found this proof http://www.apronus.com/geometry/triangle.htm . I have some more questions though:

Who came up with this proof?
What's the proof for <BAC = < B'CA
As a mathematician when do you give up trying to find such proofs for yourself (assuming you fail to) and just look up the answer? --Polyknot (talk) 03:35, 12 October 2017 (UTC)[reply]

I have no idea who was the first to come up with the proof, but the first known to publish the theorem was Euclid – it was Proposition 32 in the Book I of his Elements.^[1] --CiaPan (talk) 06:55, 12 October 2017 (UTC)[reply]

As to the question about proving <BAC = < B'CA, this is Euclid's Proposition 29 of book I of the Elements.^[2] The proof depends on the Parallel Postulate. --69.159.60.147 (talk)

References

^ David E. Joyce (1996). "Euclid's Elements, Book I, Proposition 32".
^ David E. Joyce (1996). "Euclid's Elements, Book I, Proposition 29".

Smith set and Smith-losing set

In elections where the Smith set doesn't contain all candidates, can it contain a candidate who'd still be in it if all preference rankings were reversed? Neon Merlin 06:30, 12 October 2017 (UTC)[reply]

Huh... no? If A is a candidate from outside the Smith set (A exists per the original assumption), and we reverse all preferences, then A beats any candidate from the old Smith set in the new preference order, and thus none of the "old Smiths" are in the new Smith set. Tigraan^{Click here to contact me} 08:57, 12 October 2017 (UTC)[reply]

Tigraan, I think there's a non sequitur in your final clause ("and thus..."). Just because A defeats B which was in the old Smith set, that doesn't mean that B is not in the new Smith set. If the new Smith set contains at least two elements, then assuming no ties, it contains an element that is defeated by another element. Loraof (talk) 18:58, 12 October 2017 (UTC)[reply]

Hmm, I missed that. But I am still right

, because new Smiths cannot be old Smiths.

I actually pieced together a ridiculously long proof (which I will leave below in case the revised one has a fault). However: the complementary of the old Smith set is a dominant set (each element beats each non-element) in the new order (and it is non-empty by assumption). It may not be minimal, but it proves the intersection between new and old Smiths is empty, which ~~patches~~ replaces the above demo. Tigraan^{Click here to contact me} 12:40, 13 October 2017 (UTC)[reply]

Ridiculously long proof of a useless lemma - do not read

Lemma: if an element from the old Smith set is in the new Smith set, then the new Smith set contains the whole set.

Let S be the whole set of candidates. Let S_1 be the Smith set in the original order, i.e. the minimal set such that $\forall a\in S_{1},\forall b\notin S_{1},a>b$ . Let S2 be the new Smith set, and $A=S_{1}\cap S_{2}$ . Assume that A is nonempty, and let a be one of its elements.

In that case, $\forall b\notin S_{1},b\in S_{2}$ because each of those b beats a in the new order. Then there are two cases depending on how much of S1 is included in S2:

If A=S1, then it means S1 is included in S2. Since by the above its complementary is also included, $S_{2}=S$ .
Otherwise, $S_{1}-A$ is nonempty. It must have at least one element x that is beaten by at least one element of A in the old order, otherwise, it contradicts the minimality of S1 as a Smith set (since A would do). Thus, in the new order, x beats an element of the Smith set (since A is included in S2), and thus x is in S2. This is a contradiction per the construction of A. Hence, this case does not happen. QED

Calculator problem

If you use a traditional, non-scientific calculator and type

a + + b, then keep pressing =, the values will consistently add b.

But if you type:

a * * b, then keep pressing =, the values will consistently multiply by a.

Any reason for this inconsistency?? Please try both of these with different calculators. Georgia guy (talk) 15:18, 12 October 2017 (UTC)[reply]

I just tried it, and can confirm it on two different models of (cheapo, dollar store) calculators. 3+4 = = = = returns the pattern 7, 11, 15, 19... 3x4 = = = = returns the pattern 12, 36, 108, 324. I have absolutely no idea why. --Jayron 32 16:02, 12 October 2017 (UTC)[reply]

Georgia guy, on your calculator is it necessary to press "+" and "*" twice (as you typed), or does the same thing happen if you only press them once? I see some suggestion online that on some Casio calculators, pressing the operation key twice in succession is necessary to enter "K" mode, to avoid an inadvertent error from accidentally pressing only the "=" twice. -- ToE 18:55, 12 October 2017 (UTC)[reply]

The popup calculator that comes with Windows 7 does these repeated calculation without a double press of the operation key, and unlike your and Jayron's calculator, applies b repeatedly for both addition and multiplication. Thus 3+4=== gives 7, 11, 15 and 3*4=== gives 12, 48, 192. -- ToE 19:17, 12 October 2017 (UTC)[reply]

This YouTube video from Casio suggests that their K or Constant mode, entered when you press the operation key twice, repeatedly applies a. That is 3++4=== will give 7, 10, 13 and 3**4=== will give 12, 36, 108. This works with the inverse operations also, so 3--4=== gives 1, -2, -5 and 3÷÷4=== gives 1.3333, 0.4444, 0.1481. Thus 3--4= and 3÷÷4= will give different values than 3-4= and 3÷4=. -- ToE 19:34, 12 October 2017 (UTC)[reply]

Yes, that's how Casio scientific calculators have always worked. It's very useful for "subtracted from" and "divided into", and for VAT calculations, especially when it was 17.5% (1.175** n= then subsequent amounts= etc.) Dbfirs 15:36, 14 October 2017 (UTC)[reply]

On my calculator, double and single operator gives the same result and doesn't reverse the operands. 8+2=== gives 10, 12, 14. 8-2=== gives 6, 4, 2. 8÷2 gives 4, 2, 1. But 8*2=== gives 16, 128, 1024. I suspect it's chosen because a*b is more often^{[citation needed]} viewed as "multiplying a by something" than "multiplying something by b", while "a+b" is viewed as "adding b to something" and not "adding a to something". PrimeHunter (talk) 23:42, 12 October 2017 (UTC)[reply]

October 13

Differential equations

I had a few problems that I was stuck on:

1. Suppose a cell is suspended in a solution containing a solute of constant concentration

C s

. Suppose further that the cell has constant volume

V

and that the area of its permeable membrane is the constant

A

. By Fick's law the rate of change of its mass

m

is directly proportional to the area

A

and the difference

C s - C (t)

, where

C (t)

is the concentration of the solute inside the cell at time

t

. Find

C (t)

if

m = V \cdot C (t)

and

C (0) = C 0

. Use

k > 0

as the proportionality constant.

C (t) =

2a. According to Stefan's law of radiation the absolute temperature T of a body cooling in a medium at constant absolute temperature

T m

is given by

.mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠dT/dt⁠ = k(T4 - Tm4)

, where

k

is a constant. Stefan's law can be used over a greater temperature range than Newton's law of cooling. Solve the differential equation.

T (t) =

2b. Using the binomial series, expand the right side of the following equation. (Only write the first three terms of the expansion.)

⁠ dT / dt ⁠ = k (T 4 - T m 4)

⁠ dT / dt ⁠ = k [T m + (T - T m)] 4 - T m 4

⁠ dT / dt ⁠ = kT m 4 [1 + (⁠ T / T m ⁠)) 4 - 1]

⁠ dT / dt ⁠ = kT m 4 [(____________________) - 1]

3. A classical problem in the calculus of variations is to find the shape of a curve

C

such that a bead, under the influence of gravity, will slide from point

A (0, 0)

to point

B (x 1, y 1)

in the least time. It can be shown that a nonlinear differential equation for the shape

y (x)

of the path is

y (1 + (⁠ dy / dx ⁠) 2) = k

, where

k

is a constant. First solve for

dx

in terms of

y

and

dy

. Then use the substitution

y = k sin 2 (θ)

to obtain a parametric form of the solution. The curve

C

turns out to be a cycloid.

x (θ) =

For 1, I wrote $C (t) = ⁠ m / V ⁠$ , but didn't know how to proceed from there.

For 2a, I tried separation of variables then factoring then partial fractions:

Work

⁠ dT / dt ⁠ = k (T 4 - T m 4)

⁠ dT / T 4 - T m 4 ⁠ = k dt

⁠ dT / (T 2 + T m 2)(T + T m)(T - T m) ⁠ = k dt

⁠ A / (T 2 + T m 2) ⁠ + ⁠ B / (T + T m) ⁠ + ⁠ C / (T - T m) ⁠ = 1

, where

A, B, C \in ℝ

⁠ A (T + T m)(T - T m) / (T 2 + T m 2)(T + T m)(T - T m) ⁠ + ⁠ B (T 2 + T m 2)(T - T m) / (T 2 + T m 2)(T + T m)(T - T m) ⁠ + ⁠ C (T 2 + T m 2)(T + T m) / (T 2 + T m 2)(T + T m)(T - T m) ⁠ = 1

⁠ A (T + T m)(T - T m) + B (T 2 + T m 2)(T - T m) + C (T 2 + T m 2)(T + T m) / T 4 - T m 4 ⁠ = 1

⁠ B (T 2 + T m 2)(T - T m) / (T 2 + T m 2)(T + T m)(T - T m) ⁠ = 1

, setting

T = - T m

B (2 T m 2)(-2 T m) = 1

$B = ⁠ -1 / 4 ⁠ T m 3$

⁠ C (T 2 + T m 2)(T + T m) / (T 2 + T m 2)(T + T m)(T - T m) ⁠ = 1

, setting

T = T m

C (2 T m 2)(2 T m) = 1

$C = ⁠ 1 / 4 ⁠ T m 3$

⁠ A (T + T m)(T - T m) + B (T 2 + T m 2)(T - T m) + C (T 2 + T m 2)(T + T m) / T 4 - T m 4 ⁠ = 1

A (T + T m)(T - T m) + (⁠ -1 / 4 ⁠ T m 3)(T 2 + T m 2)(T - T m) + (⁠ 1 / 4 ⁠ T m 3)(T 2 + T m 2)(T + T m) = T 4 - T m 4

A (T 2 - T m 2) + ⁠ -1 / 4 ⁠ T m 3 (T 3 - T 2 T m + TT m 2 - T m 3) + ⁠ 1 / 4 ⁠ T m 3 (T 3 + T 2 T m + TT m 2 + T m 3) = T 4 - T m 4

A (T 2 - T m 2) + ⁠ -1 / 4 ⁠ T m 3 (- T 2 T m - T m 3) + ⁠ 1 / 4 ⁠ T m 3 (T 2 T m + T m 3) = T 4 - T m 4

A (T 2 - T m 2) + ⁠ 1 / 2 ⁠ T m 3 (T 2 T m + T m 3) = T 4 - T m 4

A (T 2 - T m 2) = T 4 - T m 4 - ⁠ 1 / 2 ⁠ T m 3 (T 2 T m + T m 3)

A = ⁠ T 4 - T m 4 - ⁠ 1 / 2 ⁠ T 2 T m 4 - ⁠ 1 / 2 ⁠ T m 6 / T 2 - T m 2 ⁠

A = ⁠ - T m 4 - ⁠ 1 / 2 ⁠ T m 6 / - T m 2 ⁠

, setting

T = 0

$A = T m 2 + ⁠ 1 / 2 ⁠ T m 4$

$A = T m 2 + ⁠ 1 / 2 ⁠ T m 4$ , $B = ⁠ -1 / 4 ⁠ T m 3$ , $C = ⁠ 1 / 4 ⁠ T m 3$

⁠ T m 2 + ⁠ 1 / 2 ⁠ T m 4 / T 2 + T m 2 ⁠ - ⁠ ⁠ 1 / 4 ⁠ T m 3 / T + T m ⁠ + ⁠ ⁠ 1 / 4 ⁠ T m 3 / T - T m ⁠ = k dt

\int ⁠ T m 2 + ⁠ 1 / 2 ⁠ T m 4 / T 2 + T m 2 ⁠ dT - \int ⁠ ⁠ 1 / 4 ⁠ T m 3 / T + T m ⁠ dT + \int ⁠ ⁠ 1 / 4 ⁠ T m 3 / T - T m ⁠ dT = k dt

(T m 2 + ⁠ 1 / 2 ⁠ T m 4) \int ⁠ 1 / T 2 + T m 2 ⁠ dT - ⁠ 1 / 4 ⁠ T m 3 \int ⁠ 1 / T + T m ⁠ dT + ⁠ 1 / 4 ⁠ T m 3 \int ⁠ 1 / T - T m ⁠ dT = k dt

(T m 2 + ⁠ 1 / 2 ⁠ T m 4) ⁠ arctan(⁠ T / T m ⁠) / T m ⁠ - ⁠ 1 / 4 ⁠ T m 3 ln(T + T m) + ⁠ 1 / 4 ⁠ T m 3 ln(T - T m) = kt + D

, where

D \in ℝ

⁠ T m 2 + ⁠ 1 / 2 ⁠ T m 4 / T m ⁠ arctan(⁠ T / T m ⁠) + ⁠ 1 / 4 ⁠ T m 3 (ln(T - T m) - ln(T + T m)) = kt + D

⁠ T m 2 + ⁠ 1 / 2 ⁠ T m 4 / T m ⁠ arctan(⁠ T / T m ⁠) + ⁠ 1 / 4 ⁠ T m 3 ln(⁠ T - T m / T + T m ⁠) = kt + D

⁠ T m 2 + ⁠ 1 / 2 ⁠ T m 4 / T m ⁠ arctan(⁠ T / T m ⁠) + ln(⁠ T - T m / T + T m ⁠) ⁠ 1 / 4 ⁠ T m 3 = kt + D

however, this answer was verified to be incorrect. I think I might have the wrong values for $A$ , $B$ , and $C$ .

For 2b, I tried $⁠ T 4 / T m 4 ⁠ + ⁠ 4 T 3 / T m 3 ⁠ + ⁠ 6 T 2 / T m 2 ⁠$ and $⁠ T 4 / T m 4 ⁠ - ⁠ 4 T 3 / T m 3 ⁠ + ⁠ 6 T 2 / T m 2 ⁠$ , both of which were verified to be incorrect.

Any help would be appreciated. 147.126.10.148 (talk) 10:23, 13 October 2017 (UTC)[reply]

Differential equation 2a.

-{\frac {dT}{dt}}=k(T^{4}-T_{m}^{4})

Note the minus sign. The body is cooling when it is warmer than the surroundings. Choose your units of time and temperature such that the equation takes the form

-{\frac {dT}{dt}}=T^{4}-1

.

Use perturbation.

-{\frac {dT}{dt}}=T^{4}-\epsilon

T(t,\epsilon )\approx \sum _{n=0}^{\infty }T_{n}(t)\epsilon ^{n}

(as

\epsilon \rightarrow 0

)

The first term in the perturbation series

T_{0}(t)=\lim _{\epsilon \rightarrow 0}T(t,\epsilon )

satisfies the differential equation

-{\frac {dT_{0}}{dt}}=T_{0}^{4}

dt=-T_{0}^{-4}dT_{0}

Integrate

t=3^{-1}T_{0}^{-3}+t_{0}

Let the beginning of time be $t_{0}=0$ .

t=3^{-1}T_{0}^{-3}

T_{0}=(3t)^{-3^{-1}}

This is the cooling formula when the surrounding is at absolute zero.

Differentiate

dT_{0}=d((3t)^{-3^{-1}})=3^{-3^{-1}}(-3^{-1})t^{-3^{-1}-1}dt=-3^{-3^{-1}4}t^{-3^{-1}4}dt

=-(3^{-3^{-1}}t^{-3^{-1}})^{4}dt=-T_{0}^{4}dt

checking that the differential equation $dT_{0}=-T_{0}^{4}dt$ is satisfied.

Insert the next term in the perturbation series

T\approx T_{0}+T_{1}\epsilon

(as

\epsilon \rightarrow 0

)

into the differential equation

-{\frac {d(T_{0}+T_{1}\epsilon )}{dt}}=(T_{0}+T_{1}\epsilon )^{4}-\epsilon

and expand to the first power in $\epsilon$

-{\frac {dT_{0}}{dt}}-\epsilon {\frac {dT_{1}}{dt}}=T_{0}^{4}+4T_{0}^{3}T_{1}\epsilon -\epsilon

The constant terms vanish and the rest is divided by $\epsilon$

-{\frac {dT_{1}}{dt}}=4T_{0}^{3}T_{1}-1

Inserting $T_{0}^{3}=((3t)^{-3^{-1}})^{3}=(3t)^{-1}=3^{-1}t^{-1}$

-{\frac {dT_{1}}{dt}}={\frac {4T_{1}}{3t}}-1

or

({\frac {d}{dt}}+{\frac {4}{3t}})T_{1}=1

This is an inhomogenous Linear differential equation#First-order equation with variable coefficients.

The integrating factor is

e^{\int {\frac {4dt}{3t}}}=e^{{\frac {4}{3}}\ln t}=t^{\frac {4}{3}}

and the differential equation becomes

d(T_{1}t^{\frac {4}{3}})=t^{\frac {4}{3}}dt

integrate

T_{1}t^{\frac {4}{3}}={\frac {3}{7}}t^{\frac {7}{3}}+C

and divide

T_{1}={\frac {3}{7}}t+Ct^{-{\frac {4}{3}}}

Bo Jacoby (talk) 12:24, 13 October 2017 (UTC).[reply]

(Bo, I don't know what you're doing here, but it's probably well beyond the scope of what's expected.) The original poster's attempt at separating variables, and then integrating via partial fraction decomposition is most likely what's intended. I'll point out that in general, when you have a term like

{\frac {1}{T^{2}+T_{m}^{2}}},

the numerator needs to be

AT+B,

not just

B.

In this case, it happens to work out because

A=0,

but it won't always be like that. Otherwise, it's just a matter of being more careful with your algebra, because I think you've got the basic idea right (and I'm not going to try to wade through all your work, sorry

). --Deacon Vorbis (talk) 15:35, 14 October 2017 (UTC)[reply]

Thank you! I expect that the differential equation is not solvable in terms of elementary functions. Most differential equations are not. So I tried to find an asymptotic expression. I think I failed. As

t\rightarrow \infty

you should get

T\rightarrow 1

, and that is not what I got. I must have made mistakes. Bo Jacoby (talk) 18:51, 14 October 2017 (UTC).[reply]

Summary. The equation is

-{\frac {dT}{dt}}=k(T^{4}-T_{m}^{4})

. Substituting

T=T_{m}y

gives

3dy=-3kT_{m}^{3}(y^{4}-1)dt

. Substituting

x=3kT_{m}^{3}t

gives the dimensionless differential equation

3dy+y^{4}dx=dx

.

This is attacked by perturbation theory:

y\approx y_{0}+y_{1}\epsilon

satisfies

3dy+y^{4}dx=\epsilon dx

for

\epsilon \rightarrow 0

.

The unperturbated equation is

3dy_{0}+y_{0}^{4}dx=0

. The solution is

y_{0}=x^{-3^{-1}}

.

The perturbated equation

3d(y_{0}+y_{1}\epsilon )+(y_{0}+y_{1}\epsilon )^{4}dx=\epsilon dx

gives

3dy_{1}+4^{1}3^{-1}x^{-1}y_{1}dx=dx

. The solutions are

y_{1}=7^{-1}x+Cx^{-3^{-1}4}

where

C

is a constant of integration.

The approximate solutions to the equation

3dy+y^{4}dx=dx

are

y\approx x^{-3^{-1}}+7^{-1}x+Cx^{-3^{-1}4}

.

Bo Jacoby (talk) 06:42, 15 October 2017 (UTC).[reply]

October 14

Basic math, compound interest, multiply out

In the common compound interest formula: Capital C, interest rate i and periods p:

C * (1 + i)^p

multiplying out would lead to:

(C + iC)^p

which is a completely different result, so the parenthesis has to be solved first. But isn't algebraically valid transforming:

(ab+ac)^2

into

a(b+c)^2

or back? Shouldn't the rule rather be expressed as:

C * ((1 + i)^p)?

--Dikipewia (talk) 12:46, 14 October 2017 (UTC)[reply]

No.

(ab+ac)^{2}=(a(b+c))^{2}=a^{2}(b+c)^{2}\neq a(b+c)^{2}.

And the extra set of parentheses aren't necessary (but aren't wrong either) because exponentiation is generally understood to have higher precedence than multiplication. --Deacon Vorbis (talk) 12:57, 14 October 2017 (UTC)[reply]

You cannot get from

C * (1 + i)^p

to

(C + iC)^p

because power (Exponentiation) comes first then multiplication. 110.22.20.252 (talk) 15:04, 14 October 2017 (UTC)[reply]

Another way to prove that (ab+ac)^2 ≠ a(b+c)^2, is to show one counterexample, so let's use a=2,b=3,c=4:

(ab+ac)^2         ≠ a(b+c)^2
((2)(3)+(2)(4))^2 ≠ 2(3+4)^2
(6 + 8)^2         ≠ 2(7)^2
(14)^2            ≠ (2)(49)
 196              ≠ 98

StuRat (talk) 18:06, 15 October 2017 (UTC)[reply]

Dikipewia, you'd need to have the same exponent on both sides. For example: C^p * (1 + i)^p = (C + iC)^p. Alternatively, it is also true that C * (1 + i)^p = [C^(1/p) + i * C^(1/p)]^p, but as C is not a dimensionless quantity, the practical value of this equation is pretty much nil. 78.0.216.92 (talk) 03:05, 16 October 2017 (UTC)[reply]

October 15

Rational trigonometry

The mathematician Norman J. Wildberger writes this lemma here (at 04:48), but does not prove it. Can anyone help? יהודה שמחה ולדמן (talk) 17:56, 15 October 2017 (UTC)[reply]

Let the sides of Q be x and y. Then the area of Q is xy, and the area of Q₁ and Q₂ are x² and y². Multiply everything up (remember (xy)² = x² y²) and you get the formula in the lemma. Staecker (talk) 18:14, 15 October 2017 (UTC)[reply]

Now it occurs to me that Wildberger is trying to argue everything without ever using distances. In that case I'm not sure how to prove the lemma. Staecker (talk) 19:35, 15 October 2017 (UTC)[reply]

My advice is to ignore him. He has some rather cranky views, and it's probably not worth your time. --Deacon Vorbis (talk) 21:37, 15 October 2017 (UTC)[reply]

Wildberger is one of the few mathematicians who appreciates that mathematics has been contaminated by pseudoscientific notions about infinity. The problem is not that these notions can't be treated rigorously within mathematics, they can be so it's not that the math is wrong, rather that it becomes unnecessarily convoluted. This doesn't mean that Wildberger's approach is optimal, but it's worthwhile to think about these issues instead of treating everyone who is an Atheist w.r.t. the God of Infinity as an outcast. Count Iblis (talk) 22:43, 15 October 2017 (UTC)[reply]

Math has no scientific notions, psuedo- or otherwise. And to quite the contrary, Wildberger argues precisely that infinite sets cannot be treated rigorously; he claims that logical contradictions result from the axioms of set theory. However, when pressed for details, at best he'll evade the question, or at worst (especially when talking to someone who doesn't have any formal training in math), he'll try to dazzle with dishonest ~~bullshit~~ nonsense. This is what makes him a crank. If he wants to reject infinite sets on aesthetic grounds and develop stuff without them, then he's more than welcome to, even if the vast majority quietly ignore him. But his continual claims that everyone else is wrong—that's why he's a crank. And also, your religious metaphors here are completely inappropriate. --Deacon Vorbis (talk) 23:05, 15 October 2017 (UTC)[reply]

From an abstract point of view, math is totally isolated from science. However, there is a reason why mathematicians have chosen to work within some framework. Why were real numbers defined in the way they actually were? There are intuitive notions behind this that came for classical physics as practiced in the 19th century. If math were totally disconnected from science, it would have been a lot more cumbersome to use math in science. Now, the modern view in physics is that the continuum should be understood as a limit, it doesn't exist a priori in a physical structure, see e.g. the third paragraph on page 12 here. Had this been known back in the 19th century then it's likely that topics like analysis would have been set up in a different way involving the modern notions of scaling and renormalization as we do in physics.

Now, Wildeberger is talking to a lay audience on YouTube, he has to sell his ideas there and that's not going to be done successfully if he were to stick to the orthodox way of doing things, no one would watch his videos. In this day and age, you need to be provocative or else you'll not be heard (which is why Donald Trump is president instead of a more reasonable person). Count Iblis (talk) 00:43, 16 October 2017 (UTC)[reply]

For some reason, I seem to recall that 't Hooft is a bit of a fringe guy himself, although I don't really know this stuff. (I did notice in that paragraph that he's using "dense" incorrectly though, at least from a mathematical terminology standpoint). But I don't really know what your point is. Wildberger isn't even happy with rational numbers, or the naturals, or any infinite sets, let alone the reals. And like I said above, his objections run a lot deeper than just aesthetic or utilitarian ones. And being provocative doesn't excuse being dishonest, which I've encountered from him personally while trying to engage him in the comments on those very videos. Anyway, this is getting farther and farther off-topic, and I suggest we just stop here. --Deacon Vorbis (talk) 01:40, 16 October 2017 (UTC)[reply]

I've heard invective against the reals, the rationals, the negative integers, and zero, but never against the natural numbers, and I admit I'm tempted to watch one of his videos just to see what objection he could possibly have. OldTimeNESter (talk) 03:06, 16 October 2017 (UTC)[reply]

See here. Count Iblis (talk) 03:15, 16 October 2017 (UTC)[reply]

I have already noticed a while ago that the guy is just weird.

And even so, what if the area of a quadrance is irrational? How would he solve that? יהודה שמחה ולדמן (talk) 04:28, 16 October 2017 (UTC)[reply]

See rational trigonometry. Count Iblis (talk) 08:37, 16 October 2017 (UTC)[reply]

I've seen some of his foundations videos and I'd say that while his pov is not mainstream, it is interesting. Not only does he refuse to accept the concept of infinity, but he doesn't even accept very large numbers. It does seem reasonable to ask whether the smallest prime greater than π⋅10^10¹⁰⁰ is actually a number; it could never be written down as a decimal because the universe too small to contain it, even if it were possible to actually compute within the lifetime of the Milky Way. But while he does raise some interesting points, he's often dismissive of other points of view, which, as Deacon Vorbis was saying above, is the kind of thing that takes his ideas out of the realm of serious philosophical discussion. My main objection (in case anyone cares) is that at some point you have to just declare that you've done due diligence on the philosophical issues, just get on with with the math, and start delivering results that someone outside the field might find interesting and useful. You can argue about whether real numbers exist but I doubt that many physicists think the fine-structure constant isn't one. --RDBury (talk) 09:44, 16 October 2017 (UTC)[reply]

I wonder whether he thinks the fine structure constant exists as a number in mathematics. Maybe it can't be calculated using an algorithm. Do we then have physical numbers that can be measured and mathematical numbers that have been computed. Then why would we be entitled to multiply by it etc. etc., lots of stuff to get us invited on a tour. Dmcq (talk) 12:53, 16 October 2017 (UTC)[reply]

OK, that is enough... I totally lost you guys. So does he fit the bill of a crank or not? יהודה שמחה ולדמן (talk) 20:19, 16 October 2017 (UTC)[reply]

Crank. E Eng 22:47, 16 October 2017 (UTC)[reply]

The precise way things work in physics requires one to regularize things anyway, you have to impose a cut-off for fluctuations below some length scale which makes the theory de-facto discrete, you can in fact define the model to be a lattice model (but that's not a mathematical convenient way to go about things). Physical observables are obtained as the scaling limit of such a theory, the cut off is eliminated by an elaborate limit process. Classical physics has to be understood as something that arises in such a limit. But back in the 19th century the notion of a continuum that ultimately comes from the physical world came without this prescription that it is actually a result of a limit procedure, because quantum field theory was not invented yet. Mathematicians then found a way to deal with the continuum directly, but this is perhaps not the best way to go about things.

Since we live in a digital world today, we deal with the fact that smooth looking pictures are in fact made out of pixels. Children who play with their smartphones all the time, know very well that if you magnify a picture a lot you're going to see a block-like picture. There is then no reason to introduce the problematic concept of a real continuum to teach calculus. We can just as well introduce calculus based on a notion of coarse graining and scaling limits applied to a discrete system, where you get smooth functions in some appropriate scaling limit. Count Iblis (talk) 20:56, 16 October 2017 (UTC)[reply]

A child looking at a newspaper wirephoto in 1935 knew the same thing. The poor OP! E Eng 22:47, 16 October 2017 (UTC)[reply]

First, that's nonsense. Students generally don't see a rigorous characterization or construction of the reals before they learn Calculus anyway. But without them, you don't have things like the Intermediate value theorem, or the Mean value theorem, or any number of other fundamental results. You can still get something similar, but trying to explain it is going to be more complicated, and there's no good justification for all that extra complication, especially when it's not important for the basics. There's also nothing "problematic" about the reals; they work just fine. And second, this again has nothing to do with Wildberger's cranky ultrafinitist insistence that everyone else is wrong. --Deacon Vorbis (talk) 23:27, 16 October 2017 (UTC)[reply]

Every year or so I drop in to see if the Ref Desk is still as wacky as I remember. It is. I'm unwatching; see you in a year or so. DV, why do you waste your time here? E Eng 23:37, 16 October 2017 (UTC)[reply]

Reals work fine, but then anything we can every do in math is only ever going to involve a finite number of manipulations using a finite number of symbols. So, we can give an interpretation to what the symbols are supposed to mean, but the real beef is in the rules for manipulating the symbols, so ultimately it is all just discrete math anyway. Count Iblis (talk) 01:36, 17 October 2017 (UTC)[reply]

Dammit, I forgot to unwatch. I won't make that mistake again. E Eng 01:38, 17 October 2017 (UTC)[reply]

Iblis, there are some arguments to be made for ultrafinitism, but not that one. That one is complete and utter nonsense. The manipulation of symbols is one thing; what the symbols refer to is another thing completely. ---Trovatore (talk) 01:53, 17 October 2017 (UTC)[reply]

Going along with the pixel analogy - why do we bother with SVG then? They are, after all, only infinite "approximations" of their renderings on finite pixelated screens... 93.139.2.24 (talk) 01:58, 17 October 2017 (UTC)[reply]

October 17

Absolute vs. Relative units

This is mostly just a question about nomenclature.

Consider temperatures. Everybody knows that to convert Celsius to Fahrenheit, you multiply by 9/5 and add 32. But that's not always correct: if today was 5 degrees hotter than yesterday in Celsius, that doesn't mean it was 41 Fahrenheit degrees hotter, it was only 9 degrees hotter. It's as if there are two different kinds of temperatures, absolute and relative. The difference between two absolute temperature is a relative temperature. To convert relative Celsius to relative Fahrenheit, you multiply by 9/5 and don't add 32. So, furthermore, it's as if these two different kinds of temperatures, absolute and relative, also have different sets of units. (And, yes, speaking of "absolute temperatures", I do know about Kelvin and Rankine, but I'm overlooking them for the moment.)

Or, consider date/timestamps. Again, these come in absolute and relative flavors. The difference between two absolute date/timestamps is a relative time. And, again, there's something different about the units. The difference between January 1, 2017 at 12:00 and November 29, 2016 at 11:00 is 33 days and one hour, or 793 hours, or 47580 minutes, or maybe 2854800 seconds. The calculation isn't trivial: to come up with that 33 day number, we had to know that there are 30 days in November and 31 days in December. Instead of saying "33 days", we might have said "one month and two days", although in that case it's more than a little bit difficult to decide how many days there are in a relative month, that is, whether it should have been "one month and two days" or "one month and three days" or something else. (And to make things worse, there was a leap second in there, too, so the 2854800 number is suspect as well.)

Or, yet again, consider pointers in the C programming language. (Apologies if I'm drifting too close to RD/C territory, here.) If I have two pointer values, say, 0x12345abc and 0x1234bcde, I can subtract them, but the answer I get depends on circumstances. If they're byte pointers I'll likely get the obvious answer 25122 (0x6222), but if they're pointers to 4-byte integers they're scaled by the size of the pointed-to objects, so the answer will be 6280, since there are only 6280 4-byte ints encompassed by the two pointers. Finally, if they are void pointers, or if they do not both point within the same "object", strictly speaking it is not meaningful to subtract them at all. And, in any case, the units are clearly different: the difference between two pointers is not a pointer; it is an integer of type ptrdiff_t.

So my question is simply, is there a general theory of, or a well-known term for, this distinction between "absolute" and "relative" measurements? They all have the same properties: the difference between two absolutes is a relative, an absolute plus a relative is another absolute, you can add and subtract relatives all you want, but adding two absolutes is meaningless.

From a group theoretical point of view, I guess it's clear that absolute numbers are not closed under subtraction.

Steve Summit (talk) 02:45, 17 October 2017 (UTC)[reply]

P.S. Sometimes you can cheat. Old-time C programmers liked to assume that pointers were integers, meaning that you could do any kind of arithmetic on them you wanted. Unix to this day likes to represent both absolute and relative times as monotonically-increasing counts of seconds, and this almost works, except for those pesky leap seconds.)

P.P.S. In either Surely You're Joking, Mr. Feynman! or What Do You Care What Other People Think?, Richard Feynman rails against a science textbook problem that asked you to find the average temperature of several stars. His complaint was that it's an artificial, scientifically meaningless thing to do; mine is that in order to take the average of some temperatures, first you have to add them all up, which is problematic if, as I claim, addition of absolute numbers is ill-defined. This was brought home to me when I discovered that a certain unit-agnostic scientific quantity library that I work with has actual problems with this: if you try to add two temperatures, the sum ends up being significantly off. Not just by 32, but by 273.15. Where that number comes from is left as an exercise for the reader.

P.P.P.S. And if you think there's no problem adding absolute numbers, what's 32°C + 32°F? Or if that's too easy, what's January 1 + February 2? Or if that's too easy, what should it mean to say int a, b, *p = &a + &b; in C?

This has to do with UNITS. Basically in both advance calculators and CAS (Computer Assisted Symbolic), units are kind of a problem to be deal with. Basically there are two types of units, one where all units agree at zero. For example, 0 kilogram is equal to 0 pounds is equal to 0 grams. In this case the absolute unit is the same as the unit difference. Example: one absolute kilogram is the same as one kilogram difference.

In the units where all the units disagree at zero, for example: Kelvin and Celsius , you have the problem where one absolute unit is not the same as one unit difference. This also means that in any formula that involves that unit, you must specify if it is absolute unit or unit difference. 110.22.20.252 (talk) 05:39, 17 October 2017 (UTC)[reply]

[1] David E. Joyce (1996). "Euclid's Elements, Book I, Proposition 32".

[2] David E. Joyce (1996). "Euclid's Elements, Book I, Proposition 29".

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