Putnam Problem of the Day

[Anonymous No.4341557]

Show there do not exist four points in the Euclidean plane such that the pairwise distances between the points are all odd integers.

[Anonymous No.4341583]

first

[Blackman !medz1J/1Ok No.4341613]

>Read this.
>A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
>Is represented as:
>1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.
>If:
>H-A-R-D-W-O- R- K
>8+1+18+4+23+ 15+18+11 = 98%
>And:
>K-N-O-W-L-E- D-G-E
>11+14+15+23+ 12+5+4+7+ 5 = 96%
>But:
>A-T-T-I-T-U- D-E
>1+20+20+9+20+ 21+4+5 = 100%
>THEN, look how far the love of God will take you:
>L-O-V-E-O-F- G-O-D
>12+15+22+5+15+ 6+7+15+4 = 101%
>Therefore, one can conclude with mathematical certainty that:
>While Hard Work and Knowledge will get you close, and Attitude will get you there, It's the Love of God that will put you over the top!
>It's up to you if you share this with your friends & loved ones just the way I did..
>God bless you

[Anonymous No.4341632]

>>4341613
>But if you think the Love of God will get you far, just wait till you see how far sleeping will get you!
>Z-Z-Z-Z-Z-Z-Z-Z-Z-Z
>26+26+26+26+26+26+26+26+26+26 = 260%
>Get more sleep!

[Anonymous No.4341782]

I-N-C-E-S-T-U-O-U-S-S-O-D-O-M-Y

[Anonymous No.4341873]

This looks suspiciously easy, since I can make sense of it despite the fact these questions generally go well over my head. There's probably a catch somewhere.

[Anonymous No.4341907]

>>4341873
>show
there's the catch son
this thread is making me realize I've totally lost my trigonometry
fuck law school

[Anonymous No.4341930]

if a to b is odd, and b to c is odd then a to c is sqrt(odd*odd + odd*odd) = sqrt(odd+odd) = sqrt(even)

can only be even*even
thus no go

[Anonymous No.4341957]

>>4341930
No one said that the three points have to make a right triangle.
>>4341873
FLT is easy to understand, but extremely difficult to prove.

[Anonymous No.4341965]

>>4341957
if its not a right angle triangle split it into 2 right angle triangles, the line you split in half will have to be the sum of 2 odd, ala even.

[Anonymous No.4341973]

I'm pretty sure there's no four points such that every pairwise distance is an integer at all.

But how do I prove that?

[Anonymous No.4341979]

>>4341973 samefag

except on a straight line

[Anonymous No.4341981]

>>4341965
No, it won't. The height doesn't have to be odd. Here's a hint, if they ask for 4 points, then you're going to have to use 4 points.

>>4341973
Really? (0,0),(0,1),(0,2),(0,3).

[Anonymous No.4341987]

>>4341979
(0,0), (0,3), (4,0), (4,3)

[Anonymous No.4341988]

>>4341979
Or if the points are the vertices of a rectangle with sides 3 and 4 (or the two legs of any right triangle with integer sides).

[Anonymous No.4341999]

>>4341981

O + O = E

if you make height odd, then you get E + E = E

[Anonymous No.4342032]

>>4341999
No, you're not even guaranteed that it's an integer.
(0,0),(5,0),(13/2,3sqrt(35)/2) There is a triangle with sides 5,9,11.
http://www.wolframalpha.com/input/?i=point+%280%2C0%29%2C%285%2C0%29%2C%2813%2F2%2C3sqrt%2835%29%2F2
%29

[Anonymous No.4342042]

>>4341988
(0,0),(13,0), (433/26,15sqrt(35)/26), (76/13,12sqrt(35)/13)
http://www.wolframalpha.com/input/?i=++point+%280%2C0%29%2C%2813%2C0%29%2C+%28433%2F26%2C15sqrt%2835
%29%2F26%29%2C+%2876%2F13%2C12sqrt%2835%29%2F13%29

Sides are 13,5,11,8. Diagonals are 9,17.

[Anonymous No.4342052]

>>4341965
An equilateral triangle with an odd length disproves this easily. Besides if it was that easy they would ask you to prove that no triangle can have all three sides odd, which is false.

[Anonymous No.4342095]

Maybe: Suppose that 5 of the 6 pairs are odd integer lengths. Prove that the last pair cannot have an odd length.

[Anonymous No.4342121]

Well obviously you start with four arbitrary points (put one of them at the origin to simplify calculations)
and then I think if you use

k^2 = 1 mod 8, k is odd integer

then you'll find a contradiction

[Anonymous No.4342126]

Oh crap. Nowhere does it say that they have to form a quadrilateral does it?

[Anonymous No.4342130]

I dunno, i guess start with (a,b),(c,d),(e,f),(g,h) as the four points, numbers will get you know where
Distance from (a,b) to (c,d) using pythagorean theorem is (a-c)^2+(b-d)^2=j... for values k, l, and m
Do this for all points and get a crazy system equations, solve for things, mauybe ending up with something like k=2m, proving you right

I have no idea if this is the right method though, just a guess

[Anonymous No.4342134]

>>4342130
It's more convenient (and just as general) if you use
(0,0) , (0,t) , (x,y) , (a,b)
as your four points

[Anonymous No.4342163]

>>4342130
I am this person
>>4342134
Not really, you just defined one point and then gave another point the same x value, making it much more specific (though it is a lot more convenient)

[Anonymous No.4342171]

>>4342163
well, given any 2 points, you're going to form a (straight) line, so no matter how you have your points oriented you could have them rotated and translated to have the same x value

[Anonymous No.4342177]

>>4342163
No, it's not more specific at all. Any coordinate system is just as correct as the other.
I'm just positioning the axes so that two points fall along it, one at the origin. (which can be done for any two points without changing the shape or size of the figure)

[Anonymous No.4342191]

>>4342121
I had to go out to 16.

Place one point at (0,0) and another at (0,a). Let <span class="math">(x_1,y_1)[/spoiler] and Let <span class="math">(x_2,y_2)[/spoiler] denote the coordinates of the remaining points.

<span class="math">x_1^2 + y_1^2 = b^2[/spoiler]
<span class="math">(x_1-a)^2 + y_1^2 = c^2[/spoiler]
<span class="math">2a x_1 - a^2 = b^2 - c^2[/spoiler]
<span class="math">x_1 = \displaystyle{a^2 + b^2 - c^2 \over 2a}[/spoiler]
<span class="math">y_1 = \displaystyle{\sqrt{4a^2 b^2 - (a^2 + b^2 - c^2)^2} \over 2a}[/spoiler]
<span class="math">y_1 = \displaystyle{\sqrt{2a^2 b^2 + 2a^2 c^2 + 2b^2 c^2 - a^4 - b^4 - c^4} \over 2a}[/spoiler]

Similarly, letting
<span class="math">x_2^2 + y_2^2 = d^2[/spoiler]
<span class="math">(x_2-a)^2 + y_2^2 = e^2[/spoiler]
we have
<span class="math">x_2 = \displaystyle{a^2 + d^2 - e^2 \over 2a}[/spoiler]
<span class="math">y_2 = \displaystyle{\sqrt{2a^2 d^2 + 2a^2 e^2 + 2d^2 e^2 - a^4 - d^4 - e^4} \over 2a}[/spoiler]

Last, require
<span class="math">(x_1 - x_2)^2 + (y_1 - y_2)^2 = f^2[/spoiler]

For any odd number n,
<span class="math">n^2 = 1 ~ or ~ 9 \pmod{16}[/spoiler]
<span class="math">2n^2 = 2 \pmod{16}[/spoiler]
<span class="math">4n^2 = 2 \pmod{16}[/spoiler]
<span class="math">n^4 = 1 \pmod{16}[/spoiler]

So
<span class="math">2a x_1 = 1 ~ or ~ 9 \pmod{16}[/spoiler]
<span class="math">2a x_2 = 1 ~ or ~ 9 \pmod{16}[/spoiler]
<span class="math">(2a y_1)^2 = 3 \pmod{16}[/spoiler]
<span class="math">(2a y_2)^2 = 3 \pmod{16}[/spoiler]
<span class="math">(2a y_1)^2 (2a y_2)^2 = 9 \pmod{16}[/spoiler]

But
<span class="math">(2a x_1 - 2a x_2)^2 + (2a y_1 - 2a y_2)^2 = 4 a^2 f^2[/spoiler]
<span class="math">(2a x_1 - 2a x_2)^2 + (2a y_1)^2 + (2a y_2)^2 - 2(2a y_1)(2a y_2) = 4 a^2 f^2[/spoiler]
<span class="math">0 + 3 + 3 - 2(2a y_1)(2a y_2) = 4 \pmod{16}[/spoiler]
<span class="math">2(2a y_1)(2a y_2) = 2 \pmod{16}[/spoiler]
<span class="math">(2a y_1)(2a y_2) = 1 ~ or ~ 9 \pmod{16}[/spoiler]
<span class="math">(2a y_1)^2 (2a y_2)^2 = 1 \pmod{16}[/spoiler]

Somebody check this over to see if I made any mistakes again. I've found a number of erroneous proofs of this tonight.

[Anonymous No.4342203]

I'll do it using
k^2 = 1 mod 8

>>4342121
I'll start by using this coordinate system.

By the Pythagorean theorem, the squares lengths of the 6 lines for the quadrilateral are as follows:
From point (0,0):
t^2
x^2 + y^2
a^2 + b^2
From point (0,t)
x^2 + (y-t)^2
a^2 + (y-t)^2
And lastly
(x-a)^2 + (y-b)^2

We assume that all of these lengths are odd, and then seek a contradiction.

If each of these values are odd, then by the modular equality mentioned above, they are each all congruent to 1.

We can subtract the first from the second,
r^2 - (r^2-2ar+a^2) + s^2 - s^2 = 0 mod 8
2ar = a^2 mod 8

[Anonymous No.4342217]

>>4342203
What are r and s?

[Anonymous No.4342226]

>>4342191
>Place one point at (0,0) and another at (0,a)
Sorry, this should be (a,0).

[Anonymous No.4342229]

>>4342203
Assuming the coordinates got renamed somehow in the last step to
(0,0), (a,0), (r,s), (?,?)

I still don't see a contradiction here. a^2 must be congruent to 1 mod 8, but nothing has been proven about r, which need not even be an integer.

[Anonymous No.4342238]

>>4342229
Doesn't look like he finished copying the proof from
http://www-bcf.usc.edu/~lototsky/PiMuEp/Putnam1985-2000.pdf

[a8sE !/math/a8sE No.4342267]

>>4342126
lol

[Anonymous No.4342624]

As pointed out earlier, we may assume witout loss of generality, that one point is on the origin, and another on the x-axis.
Hence we have (0,0) and (0,a), where a is an odd integer. Now, it's obvious that no other point can be on the x-axis, since if it's (0,b), then, either b-0 is not an odd integer, or b-a is not an odd integer.
Hence, the third point must be of the shape (b,c). Where |(b,c),(0,0)| and |(b,c),(0,a)| are both odd integers.
Hence, \sqrt(b^2+c^2) and also b^2+c^2 is an odd integer. Similarly, b^2+(c-a)^2 is an odd integer. b^2 + c^2 - 2ac + a^2 is an odd integer. Since we know that b^2 + c^2 is an odd integer, and a^2 is an odd integer, 2ac must be an odd integer.
The same reasoning holds for the fourth point (d,e) [where b replaced by d, and c by e]. But also (d-b)^2 + (e-c)^2 should be an odd integer. Hence d^2 - 2db + b^2 + e^2 - 2ce + 2^c must be odd, and - 2db - 2ce must be an odd integer.
However, - 2ac - 2ae is even, so we've reached a contradiction.

[Anonymous No.4342709]

>>4342624
What is the contradiction?
If you showed that (-2db - 2ce) was even I would agree, but you only showed that (2ac-2ae) was even.

[Anonymous No.4342954]

What kind of formal education do you all have? Because I'm halfway through 6th form, having taken pure/further pure 1&2, and stats, mechanics, and decision 1, and this is all miles beyond what I've learned to far.

[Anonymous No.4342958]

>>4342954
What a waste of money.

[Anonymous No.4342961]

>>4342958
wat

I haven't paid for any of it, and all the classes would continue regardless of my attendance. I'm not wasting anyone any money.

[Anonymous No.4343000]

>>4342961
It's spelled "what", sweetheart.

You're wasting plenty of money. Think about it.

[Anonymous No.4343010]

>>4343000
It's a meme, son.

I'm not wasting any money.

[Anonymous No.4343017]

>>4343010
Your poor money. *tear*

[Anonymous No.4343084]

>>4342954
The maths knowledge required is not very high. It requires intelligence and problem solving capabilities. You can almost get to graduation without having this.
Some problems are easier than others, though. For instance, tomorrow (assuming we don't skip a problem) we should have
"Suppose that a sequence <span class="math">a_1,a_2,a_3,\dots[/spoiler] satisfies <span class="math">0<a_n\leq a_{2n}+a_{2n+1}[/spoiler] for all <span class="math">n\geq 1[/spoiler]. Prove that the series <span class="math">\sum_{n=1}^{\infty}a_n[/spoiler] diverges.", which is a fairly easy one.

[Anonymous No.4343092]

>>4343084
I'm intuitively good with numbers; Most of what I'm learning now I've already figured out in the past- I remember determining how to sum an arithmetic series when I was 6.

My question is definitely when, if at all, you were taught to do this

[Anonymous No.4343114]

>>4343092
All of the Putnam problems can be done by 1st or 2nd year university students. Sometimes using higher level theorems can make it easier, but that's not required. The problems are designed for a competition, and the egigibility is
>The competition is open only to regularly enrolled undergraduates, in colleges and universities of the United States and Canada, who have not yet received a college degree. No individual may participate in the competition more than four times. An eligible entrant who is also a high school student must be informed of this four time limit.
From which you can deduce that some very good high school students can participate, and that people are encouraged to start participating a few years before they graduate.

[Anonymous No.4343122]

>>4343114
Righty ho. Thanks.

[Anonymous No.4343155]

>>4343114
I think I got it. Pythagorean triangles can't have three integer length sides, because an odd number squared plus/minus an odd number squared will give an ever number.

[Anonymous No.4343174]

>>4343155
...and as an extension of that, any non-Pythagorean triangle can be filled by two with right angles, and the distance of the two sides on the original triangle's hypotenuse will have to be even, as if it was odd that would give the two smaller triangle's non-integer values for that side, meaning the hypotenuse of their own triangles couldn't be an odd integer.

[Anonymous No.4343186]

>>4343174
A triangle can have 3 odd integer sides.

5-7-9 triangles do exist.

[Anonymous No.4343191]

>>4343186
crap.

Back to the drawing board then

[Anonymous No.4343864]

I feel like a dumbass.

I looked up pairwise but still have no idea what the hell it means. Any help for a newb?

[Anonymous No.4343880]

This sounds like either an application of linear algebra or the pigeon hole principle, or both. Seeing as I'm busy reading papers though I won't be able to help.

[Anonymous No.4343882]

What the hell does pairwise mean? Google gives a bunch of answers for it, with none making sense in this context.

I feel like a dumbass.

[Anonymous No.4343907]

>>4343882
>pairwise

You're giving me that feel again ;(
>no gf

[Anonymous No.4343924]

The points don't have to be at integer co-ordinates. They can be rational, or the square root of rational numbers or some weird co-ordinates. For example, The distance between (0,0) and <span class="math"> (36 - \sqrt{1305} , \sqrt{72 \sqrt{1305}}) [/spoiler] is 51. The first step is determining what form the co-ordinates have co-ordinates have, which might be a bit complicated. Then after you write them as a function of integers, then you can go about doing your (mod whatever) stuff. That's the easy part.

[Anonymous No.4344186]

It is impossible to establish the internal logical consistency of this deductive system.

These are not proofs.

[Anonymous No.4344234]

>>4344186
>It is impossible to establish the internal logical consistency of this deductive system.
Ok

>These are not proofs.
So the previous sentence was lucky, you didn't actually understand it: understanding it would mean that you know the definition of a proof, and you would therefore know that they do depend on a system. The consistence of the system has nothing to do with the proofs in that system.

[Anonymous No.4344274]

Here's another approach that might be simpler. Using Heron's formula, that the area of a triangle with side lengths, a, b, and c is: <div class="math"> A = \sqrt{s(s-a)(s-b)(s-c)} </div> where <span class="math"> s = \frac{a+b+c}{2} [/spoiler], which can be rewritten as: <div class="math"> A = \frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{4} </div>
Then we know that the area of a triangle with all odd integer sides has an area in the form of either <span class="math"> \frac{2k+1}{4} [/spoiler], where k is an integer, or <span class="math"> \frac{\sqrt{2k+1}}{4} [/spoiler], where k is an integer, and 2k+1 is not a perfect square.
Keeping that in mind, now look at the image. The four points are either arranged with a three point convex hull, with the other point in the middle, or a four point convex hull. If the three-point hull is the case, than three smaller areas add up to the biggest one. If the four point hull is the case, then the sum of two areas, is equal to the sum of two other ones. This is about as far as I got, but maybe a smarter /sci/entist than me can pick up where I left off.

[Anonymous No.4344284]

Here's what I was talking about with the hulls.

[Anonymous No.4344533]

>>4344234

>So the previous sentence was lucky, you didn't actually understand it: understanding it would mean that you know the definition of a proof, and you would therefore know that they do depend on a system. The consistence of the system has nothing to do with the proofs in that system.

Lolwut? The logical consistency of a system has nothing to do with the proofs of that system? If math weren't logically consistent mathematical proofs would be invalid. I guess you could still call them proofs in the sense that I can call the color red blue, but for all intents and purposes they would cease to be proofs.

[Anonymous No.4344677]

>>4344533
You don't know what "consistent" mean for a logics system. You basically don't understand what Godel is about.

[Inane Trolling Garbage !Pyd2/R7DNA No.4345082]

>>4344677
Well you don't understand what sex is about.

[Anonymous No.4345109]

>>4345082
>be on a science & maths board
>insult people for talking about maths
>has a tripcode to troll
>thinks he's funny because his reference in terms of humor are the jokes he hears from his stepfather when he's getting assraped

[Inane Trolling Garbage !Pyd2/R7DNA No.4345116]

>>4345109
>>the jokes he hears from his stepfather when he's getting assraped

WELL AT LEAST I GET LAID

[Anonymous No.4345124]

>>4345116
So do I, I'm your stepfather.

[Inane Trolling Garbage !Pyd2/R7DNA No.4345129]

>>4345124
Can we use lube tonight?

[Anonymous No.4345142]

>>4345129
Let's talk about this out of /sci/, son. I'm off.

[Anonymous No.4345178]

>>4344677

Except I do and have taken more formal logic courses than you as I am currently conducting research in analytic philosophy.

Strawman.jpg

[Anonymous No.4345245]

>>4345178

Then either you learned badly or you express yourself badly.

Of course it was shown that mathematics cannot prove its own consistency.

But that has no bearing over the notion of proof. A proof just means that you can derive it from elementary axioms. The choice of the axioms is arbitrary.

[Anonymous No.4345447]

>>4342203
Sorry for the delay. My laptop ran out of battery last night, and I didn't feel like getting up to get the charger.

Anyway, the variables for the last equation are different because I wrote it out first, but used different variables to identify the points.

Resuming where I left off.

t^2
x^2 + y^2
a^2 + b^2
x^2 + (y-t)^2
a^2 + (b-t)^2
(x-a)^2 + (y-b)^2
Are all congruent to 1 mod 8.

Subtract the 2nd from the 4th
x^2 + y^2 - 2ty +t^2 - (x^2 + y^2) = 0 mod 8
t^2 = 2ty mod 8
2ty = 1 mod 8
2ty must be thus be an odd integer, and so either t or y is a rational number with an even denominator. Because t^2 = 1 mod 8, t must be an integer, and so it is y that has the even number. Additionally, in order for 2ty to be congruent to t^2, the denominator of y must divide into 2t.

The same procedure can be used to show that b also has an even denominator that divides into 2t.

Continued in next post

[Anonymous No.4345810]

[Anonymous No.4345821]

>>4345447
>>4342203
>Anyway, the variables for the last equation are different because I wrote it out first, but used different variables to identify the points.
Right, it wasn't because you just copying and failing to change variables.

[Anonymous No.4345852]

>>4345821
And here we see why half of /sci/ has given up on these threads and decided to just troll them to piss.

[Anonymous No.4345914]

Cube4 Intelligence,

SuperSymmetry Of

TimeCube4 Math Is

Absolute Proof Of

School Retardation.
No 1 Educated matches my intelligence.

Really, just how diabolic are the humans

educated on Earth based upon ONEness?

**********************************************

+1 x +1 = +1 as if a male value and

-1 x -1 = -1 as if a female opposite,

Hell awaits those who add these.

***************************

No educated person could work for me

who could not tear and burn the bible –

for he would be a liar of 1 day deception.

Humans are only using 1 of the 4 existing

days available to them. A single day is

impossible as all 4 are interdependent.
Greenwich Mean Time is wrong and evil,

for there are 4 simultaneous Days, not 1.

Greenwich has a midnight to midnight 1

corner day rotation. It has an imaginary

midday to midday with broken lines on

chart to avoid bible 1 day error conflict.

It completely ignores sunup & sundown.

Actually, Genesis 1:5 is not even 1 day.

What you have is 4 corners, no time rota.

Earth has 2 plus quads & 2 minus quads

existing as 0 as opposites but, voiding as 1.
You can’t comprehend fact that Cube4

simultaneous 24 hour days rotate within

same 24 hour rotation of Mother Earth.

[Yo, I got (tired of) this !/math/a8sE No.4345917]

>>4345447
Suppose you had succeeded in passing off someone else's proof as your own in this anonymous forum. How would you benefit?

[Anonymous No.4345967]

>>4345178
Well, then it's remarkable that you've made such an elementary mistake.
A proof is a sequence of statements that are derived from the rules of the formal system. It doesn't matter whether a formal system is inconsistent. In that case, you'll simply be able to proof anything. But it's still a proof.

[Anonymous No.4346002]

>>4342191
In setting up your coordinate system you can indeed assume WLOG that y1 > 0. The case of y1 = 0 or y2 = 0 obviously yields no solutions. But now you cannot assume that y2 > 0. It may be the negative root as well.

[Anonymous No.4346034]

>>4346002
If you have 4 points you can easily just translate them up until all of them are positive.

[Anonymous No.4346162]

>>4345178
>Except I do and have taken more formal logic courses than you
I seriously doubt that. I'm a PhD student in computer science and every time I had the choice between something theoretical and something practical, I chose the theoretical course. Also, the simple fact that you're wrong while I'm right and that you didn't even check http://en.wikipedia.org/wiki/Consistency and http://en.wikipedia.org/wiki/Completeness shows how little you question yourself and is a very bad sign for anyone in research. I wish you to change to a more mature behavior, otherwise I don't think you will be able to accomplish much.

To sum it up, as other anons have:
- You can prove any true property: the system is complete,
- There is no contradiction in the system: the system is consistent.
Here is an inconsistent system, and a proof in it:
Axiom 1: <span class="math">\vdash A[/spoiler]
Axiom 2: <span class="math">\vdash \lnot A[/spoiler]
Property to prove: <span class="math">A[/spoiler]
Proof: <span class="math">\vdash A[/spoiler]
Yet the system is inconsistent. But I doubt you understand such a MWE since you seem to be unable to understand things formally and try to philosophically interpret them ("hur Godel means proofs don't exist"...).

[Anonymous No.4346302]

>>4346162
> you didn't even check http://en.wikipedia.org/wiki/Consistency and http://en.wikipedia.org/wiki/Completeness
> I'm a PhD student in computer science

Your rotten crotch is showing. Please leave immediately.

[Anonymous No.4346337]

>>4346162
So close and yet so far.
It's ok, though, you're less wrong than the guy you're arguing against, but:
In this sense, completeness means that every statement can be proven true or false (regardless of 'actual' truth, if it even exists), and consistency means that no statement can be proven both true and false. (Sometimes completeness is used as meaning that the derivation rules only derive statements that are true according to the semantics, but not in this context.)

[Anonymous No.4346521]

>>4346302
Yet I'm a PhD student in computer science.

>>4346337
My consistency definition is a rewording of yours. Just define "contradiction" as the existence of a statement that can be proven both true and false.
My definition of completeness is indeed slightly off, it is equivalent to yours only if you assume that if you can prove <span class="math">\vdash A[/spoiler], you can prove <span class="math">\vdash \lnot \lnot A[/spoiler], and vice versa. In that case, the existence of a proof of all true properties guarantees the existence of a proof of all the negations of false properties (which I guess is what you mean by "proving a statement false"). But anyway I think my definition is correct. It's just a rewording of "Tautologies are Theorems", which is I believe the formal definition.
Anyway, that other guy tried to be a smartass with his first comment and failed, I'm still unsure why we're discussing this that far.