Wikipedia:Reference desk/Archives/Mathematics/2017 October 12

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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)

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)

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. --CiaPan (talk) 06:55, 12 October 2017 (UTC)

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

References

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? NeonMerlin 06:30, 12 October 2017 (UTC)

• 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)

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)

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)

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)

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. --Jayron32 16:02, 12 October 2017 (UTC)

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)

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)

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)

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)

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 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)