Posted: 8/31/2006 12:21 AM
Post subject: Math question
Quote
VirtualAlex
Joined: 5/31/2004
Posts: 464
Location: Minnesota
This isn't a riddle, but it's a question I have been trying to figure out for a long time but sadly I am not that good at math so I have never been able too. it's totally possible that what I am asking for is impossible also. But here it is. Is it possible to write an equation that contains 2 variables, that cannot be true? You can use numbers and variables, but exactly two variables (X and Y) must be used in this equation, they can be used multiple times each if you wish. So for example something like this: X + Y = X (Of course this can easily be true, if either X or Y are 0) It seems like making one or the other 0 will always tend to make an eqaution possible but like I said, I am bad at math.
JXQ's biggest fan.
Posted: 8/31/2006 12:26 AM
Quote
Baxter
Skilled player (1410)
Joined: 5/31/2004
Posts: 1821
TAS of 2013 SNES TAS of 2013 Speedy TAS of 2007 Innovative TAS of 2007
12Motion wrote:
X + Y = X (Of course this can easily be true, if either X or Y are 0)
This is not true if only X = 0, Y has to be 0. I'm not quite sure what you mean, but X + Y + 1 = X + Y is obviously never true.
http://tasvideos.org/Baxter.html
Posted: 8/31/2006 12:30 AM
Quote
curtmack
Former player
Joined: 6/4/2006
Posts: 97
Location: Everywhere including nowhere
That's not true Baxter... Let's take any two numbers a and b such that a=b. a=b ab=b^2 ab-a^2=b^2-a^2 a(b-a)=(b+a)(b-a) a=b+a a=2a 1=2 0=1 X+Y=X+Y+1 So there.
...?
Posted: 8/31/2006 12:34 AM
Quote
Baxter
Skilled player (1410)
Joined: 5/31/2004
Posts: 1821
TAS of 2013 SNES TAS of 2013 Speedy TAS of 2007 Innovative TAS of 2007
curtmack wrote:
a=b ab=b^2 ab-a^2=b^2-a^2 a(b-a)=(b+a)(b-a) a=b+a a=2a 1=2 0=1
Well... math doesn't allow dividing by 0. If a=b then b-a=0, and you divide by that to get a=b+a
http://tasvideos.org/Baxter.html
Posted: 8/31/2006 12:35 AM
Quote
curtmack
Former player
Joined: 6/4/2006
Posts: 97
Location: Everywhere including nowhere
Congratulations. That bit of algebra has been circulating the internet for years, actually.
...?
Posted: 8/31/2006 2:27 AM
Quote
VirtualAlex
Joined: 5/31/2004
Posts: 464
Location: Minnesota
wow how the hell did I never thing of X+Y+1 = X+Y I feel so stupid.
JXQ's biggest fan.
Posted: 8/31/2006 2:47 AM
Quote
OmnipotentEntity
He/Him
Player (36)
Joined: 9/11/2004
Posts: 2630
First we consider the identity: e^πi + 1 = 0 Which we know from Calculus, Taylor Series and all that jazz. And indeed, every formula of the form: e^(2n+1)π i + 1 = 0; where n ∈ Z But I digress. Anyway, e^πi + 1 =0 e^πi= -1 e^2πi = 1 e^2πi = e^0 2πi = 0 What's going on here?
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 8/31/2006 2:53 AM
Quote
Bag_of_Magic_Food
He/Him
Player (206)
Joined: 5/29/2004
Posts: 5712
Everything's going on here.
put yourself in my rocketpack if that poochie is one outrageous dude
Posted: 8/31/2006 1:45 PM
Post subject: Re: Math question
Quote
samurai_goroh
Editor, Player (54)
Joined: 12/25/2004
Posts: 634
Location: Aguascalientes, Mexico
12Motion wrote:
X + Y = X (Of course this can easily be true, if either X or Y are 0)
Solving 1 equation with 2 variables will have an infinite number of answers. X + Y = X X + Y - X = X - X Y = 0 Replacing Y = 0 on the original answer, you get: X + (0) = X X = X There forward X can be any value. Baxter's solution is simple. The answer of equaling 2 equations is where both equations cross (have the same value). So if you have 2 parallel lines, they will never cross each other...
I'm the best in the Universe! Remember that!
Posted: 8/31/2006 3:01 PM
Quote
anubis
Joined: 5/29/2006
Posts: 138
Baxter wrote:
X + Y + 1 = X + Y is obviously never true.
X=Y=∞ :p
Posted: 8/31/2006 5:30 PM
Quote
DK64_MASTER
Joined: 10/24/2005
Posts: 1080
Location: San Jose
What about cos(x*y) > 1 or a simple sinc(x*y) > 1 (or in fact = 1, but sinc(0) = 1 in most cases, although it should be undefined) or e^(i*pi*x*y)>1 As long as something operator has a bound, an equation with 2 variables, and 1 at least constant should be able to be false for all values of either variable, as long as your variables are real numbers.
<agill> banana banana banana terracotta pie! <Shinryuu> ho-la terracotta barba-ra anal-o~
Posted: 8/31/2006 5:45 PM
(Edited: 8/31/2006 5:51 PM)
Quote
JXQ
Experienced player (761)
Joined: 5/6/2005
Posts: 3132
NES TAS of 2008 Speedy TAS of 2008 SNES TASer of 2007 TAS of 2007 SNES TAS of 2007 Funny TAS of 2007
Is an expression using less than or greater than technically an equation? Since less than / greater than are "inequalities", I'm inclined to think relationships using them aren't equations. But if they are allowed in this question, you could just say X+Y > X+Y, which is always false. Edit: Hmm, seems I combined the two questions here into one brand new, unrelated, utterly useless question!
<Swordless> Go hug a tree, you vegetarian (I bet you really are one)
Posted: 8/31/2006 5:46 PM
Quote
OmnipotentEntity
He/Him
Player (36)
Joined: 9/11/2004
Posts: 2630
ok then, same vein then. cos(x*y) = 42
Build a man a fire, warm him for a day, Set a man on fire, warm him for the rest of his life.
Posted: 8/31/2006 5:51 PM
Quote
DK64_MASTER
Joined: 10/24/2005
Posts: 1080
Location: San Jose
OmnipotentEntity wrote:
ok then, same vein then. cos(x*y) = 42
Exactly. I used inequalities to show that I was emphasizing that bounded operators will be sufficent for this scenario. Now someone do it without using a bounded operator (i.e.your operator needs to work for any type of REAL vector space, and should not produce bounded ouput). Any volunteers? ;)
<agill> banana banana banana terracotta pie! <Shinryuu> ho-la terracotta barba-ra anal-o~
Posted: 9/2/2006 5:43 PM
Quote
VirtualAlex
Joined: 5/31/2004
Posts: 464
Location: Minnesota
You guys are very strong O_O
JXQ's biggest fan.
Posted: 9/2/2006 7:22 PM
Quote
LagDotCom
Joined: 3/7/2006
Posts: 720
Location: UK
anubis wrote:
X=Y=∞
That's no number, you foo'.
Voted NO for NO reason
Posted: 9/2/2006 9:05 PM
Quote
Dacicus
Editor, Player (69)
Joined: 6/22/2005
Posts: 1050
Would the floor and ceiling functions be what you call bounded operators? If not, you could just set one of them to a fractional value, and it would be false by definition.
Current Projects: TAS: Wizards & Warriors III.
Posted: 9/3/2006 1:14 AM
Quote
DK64_MASTER
Joined: 10/24/2005
Posts: 1080
Location: San Jose
Dacicus wrote:
Would the floor and ceiling functions be what you call bounded operators? If not, you could just set one of them to a fractional value, and it would be false by definition.
Heh, I really have no idea, but I assume they ARE bounded operators, as the domain is all reals, and the range is integers... But I'm no math major, just an EE major.
<agill> banana banana banana terracotta pie! <Shinryuu> ho-la terracotta barba-ra anal-o~