Does 2/6 Equal 1/3?

[T3chno]

You forgot to divide by the square root of pie, noob.

[Exüberance]

This is like a mathematician, physicist, and engineer joke.

Only I will use a mathematician, physicist, and computer scientist varient... and it's not a joke:

The mathematition (yes): 2/6 = 1/3 just by reducing it. No mathematician would ever have you prove something like this that is trivially true, so it can be considered true without proving it since it has already been proven by the properties of division and mathematicians are lazy. We don't write what we don't have to. (At least I don't... usually... OK sometimes I do)

The physicist (no): although 2/6 does, in fact, equal 1/3, 2/6 probably implies that 6 things were measured, and 2 of them were some case, so 2/6 is more accurate than 1/3 is, therefore, mathematically equal, but not equivalent in usefulness. You would use significant figures and error bounds, not unreduced fractions, but oh well.

The computer scientist (depends on it's data type): If we allow 2/6 and 1/3 to be floating point (or some other number type), they are equal. If, however, we create an object consisting of 2 values: numerator and denominator, that the object representing 2/6 does not equal the object representing 1/3

The layman (inconclusive): I don't care. This topic is stupid.

Quad Erat Demonstrandem (yes, I know I didn't really prove anything. I just like saying that)

[b3njamin]

Yes and No


Mathematicly seen Yes


In real life seen No (depends on situation though)


I had an old shooter game where I had 100 kills and 5 Deaths. I think it's 100/5 and not 20/1. 100/5 is harder to get and different then 20/1

[Deathconciousness]

Kablizzy is a very clever troll

he just trolled everyone in this thread that took this question seriously. How bored are you people?

[Kablizzy]

Kablizzy is a very clever troll

he just trolled everyone in this thread that took this question seriously. How bored are you people?

I just trolled everyone for FIVE STRAIGHT PAGES WITHOUT SAYING A DAMN THING.

[yungerkid]

Kablizzy is one mighty-fine troll. *whistles*

[t̷s͢uk̕a͡t͜ư]

Kablizzy is a very clever troll

he just trolled everyone in this thread that took this question seriously. How bored are you people?

I just trolled everyone for FIVE STRAIGHT PAGES WITHOUT SAYING A DAMN THING.

Did you troll everyone for 30/6 pages, or 15/3 pages?
The distinction is potentially important.

[Deathconciousness]

Kablizzy is a very clever troll

he just trolled everyone in this thread that took this question seriously. How bored are you people?

I just trolled everyone for FIVE STRAIGHT PAGES WITHOUT SAYING A DAMN THING.

let me be the one to give you a formal golf clap

*clap*

[T3chno]

Kablizzy is a very clever troll

he just trolled everyone in this thread that took this question seriously. How bored are you people?

I just trolled everyone for FIVE STRAIGHT PAGES WITHOUT SAYING A DAMN THING.

[Topo]

FIVE STRAIGHT PAGES WITHOUT SAYING A DAMN THING.

Wait, how did you make that backwards b?

[crescor]

Yes, 2/6 equals 1/3
There, I said it.

[origami_alligator]

I'm going to opt for the Suki method of solving this problem.

Which fight do you figure is more fair: A fight of 1 against 3, or a fight of 2 against 6?

Clearly, 2 against 6 is fairer. A person alone can't size up a group of opponents as well as a couple; and three people act in concert well, while six people just tend to get in each other's way and one or two would have to sit out while the first few are getting fought off.

I don't know why I was thinking of this post today, but I'm curious to know:
If the 2 people were split up to fight half of the 6 guys each, that would make the fight essentially 1 to 3, making it the unfair advantage to the 6.

So as long as you divide 2/6 in half then you get 1/3.

[t̷s͢uk̕a͡t͜ư]

I don't know why I was thinking of this post today, but I'm curious to know:
If the 2 people were split up to fight half of the 6 guys each, that would make the fight essentially 1 to 3, making it the unfair advantage to the 6.

So as long as you divide 2/6 in half then you get 1/3.

Only if the meaning you attach to the forward slash is "and" instead of mathematical division. In that case, you're saying "if you divide 2 and 6 by 2 each, you get 1 and 3" or "if you have two groups, one of 2 and the other of 6, and you divide each into two groups of the same size, then the first will become two groups of 1 and the second will become two groups of 3."
Either way, it's not mathematical division; you've redefined the meaning of that operator. Let's try to avoid this so we don't run into problems like, "if you have a rectangle whose height / width is 1/3, and you have another whose height and width is 2/6, then the second rectangle has four times the number of pennies as the first!"

[behappyy]

It kinda depends on what "Equal" means to you. If to you it means "the same as" than no it is not true. If it means "the same value" than yes.

[Exüberance]

Correct Answer to Kablizzy's Question:

In what number system?

For example, this does not hold in the number system of integers modulo 4.

[2]/[6]
=> [2] * [6]^-1 where [a]^-1 refers to the multiplicative inverse of [a]
=> [2] * [2]^-1 (6 mod 4 = 2)
however, there is no mulitplicative inverse of 2, so you cannot express 2*(6^-1) as an integer in Z[4]

But let's try [1]/[3]
=> [1] * [3]^-1
=> [1] * [3] (the multiplicative inverse of 3 in Z[4] is 3. ([3]*[3]=[9]=[5]=[1]))
=> [3]

So there you have it. In this number system, 2/6 does not exist, and 1/3 is 3.

True, I'm not using the numbers 1/3 and 2/6, but rather the number represented by 2 divided by 6 and 1 divided by 3 where division is multiplication by the mulitplicative inverse (as division is in the normal number system. x * 3 = 1, therefore x = 1/3. Same operation, Different number system. Different result)

[t̷s͢uk̕a͡t͜ư]

Correct Answer to Kablizzy's Question:

In what number system?

For example, this does not hold in the number system of integers modulo 4.

[2]/[6]
=> [2] * [6]^-1 where [a]^-1 refers to the multiplicative inverse of [a]
=> [2] * [2]^-1 (6 mod 4 = 2)
however, there is no mulitplicative inverse of 2, so you cannot express 2*(6^-1) as an integer in Z[4]

But let's try [1]/[3]
=> [1] * [3]^-1
=> [1] * [3] (the multiplicative inverse of 3 in Z[4] is 3. ([3]*[3]=[9]=[5]=[1]))
=> [3]

So there you have it. In this number system, 2/6 does not exist, and 1/3 is 3.

True, I'm not using the numbers 1/3 and 2/6, but rather the number represented by 2 divided by 6 and 1 divided by 3 where division is multiplication by the mulitplicative inverse (as division is in the normal number system. x * 3 = 1, therefore x = 1/3. Same operation, Different number system. Different result)

CORECT ANSER TO KALBIZZO'S QUESSION;
FOR WHAT DEFNITION OF SIX???????? IF U HAVE 6 IS ACTUALY 2 THEN NO WAI
SO THEIR U HAVE TI 2/6 DOES NOT EXIST BECOSE 6 IS 2 BECUASE WE'VE FOR SOME REASON DECIDED THAT IT'S APPROPRIATE TO CONSIDER A SYSTEM WHERE 6 DOESN'T EXIST FOR A PROBLEM IN WHICH 6 IS EXPLICITLY IN THE PROBLEM STATEMENT



Kablizzy: "Does 2/6 equal 1/3?"
Exuberance: "I'm going to assume, for zero reason, that you're talking about colors. Neither of those is a color, so clearly the answer is 'non-applicable'. Praise me."

[Nexx]

I liked tktktk's response best. January 3rd != February 6th (or March 1st != June 2nd).

And there are a lot of dumb posts in this thread. And some good ones. Thanks for telling me about Cantor's Diagonal Argument!

[TheRealOne]

FRACTIONS CAN END THE WORLD!!!!

a=b
Add a to both sides.
2a=b+a
subtract 2b from both sides.
2a-2b=a-b
Factor out 2 from the left side.
2(a-b)=a-b
divide both sides by (a-b).
2=(a-b)/(a-b)
Simplify the fraction.
2=1?!?!?!?!?!?!?!?!?!!? AH!!!!!!!!!!!!!!!!!!
Let your head asplode.

And according to this since 2=1, 1/3=2/3 and 2/6=1/6, and we all know that 2/3 so does no equal 1/6, therefore 1/3 does not equal 2/6.

[DemonzLunchBreak]

I hate this thread.

Also, TheRealOne has an undefined step when he divides by a-b.

[TheRealOne]

I hate this thread.

Also, TheRealOne has an undefined step when he divides by a-b.

/joke.

I was just adding more irrelevance to and irrelevant thread.

[t̷s͢uk̕a͡t͜ư]

I hate this thread.

Also, TheRealOne has an undefined step when he divides by a-b.

Try this one, then:

let a and b be any real numbers
let t = a + b

t = a + b
(a - b) t = (a - b) (a + b)
at - bt = a^2 - b^2
a^2 - at = b^2 - bt
a^2 - at + (t/2)^2 = b^2 - bt + (t/2)^2
(a - t/2)^2 = (b - t/2)^2
Two paths from here, since taking a square root gives two answers...
Taking the negative square root:
a - t/2 = -(b - t/2)
a - t/2 = t/2 - b
a + b = t
...this merely confirms our given, meaning that we didn't do anything wrong in the previous steps.
Taking the positive square root:
a - t/2 = b - t/2
a = b
QED, given any real numbers a and b, a = b. Therefore, all numbers are actually all the same number, and are just different names for that one number.

Let's try this with, to make an especially meaningful case, a = 0 and b = 1:
let t = a + b, so t = 1
t = a + b ==> 1 = 0 + 1
(a - b) t = (a - b) (a + b) ==> (0 - 1) (1) = (0 - 1) (0 + 1); -1 = -1
at - bt = a^2 - b^2 ==> (0)(1) - (1)(1) = (0)^2 - (1)^2; -1 = -1
a^2 - at = b^2 - bt ==> (0)^2 - (0)(1) = (1)^2 - (1)(1); 0 = 0
a^2 - at + (t/2)^2 = b^2 - bt + (t/2)^2 ==> (0)^2 - (0)(1) + (1/2)^2 = (1)^2 - (1)(1) + (1/2)^2; 1/4 = 1/4
(a - t/2)^2 = (b - t/2)^2 ==> (0 - 1/2)^2 = (1 - 1/2)^2; 1/4 = 1/4
So y'see, we've made completely valid statements so far. Absolutely nothing wrong with them.
But just to be sure, let's take the negative square root to verify this:
0 - 1/2 = -(1 - 1/2) ==> -1/2 = -(1/2)
Yep, negative one-half certainly equals negative one-half.
So let's move on to what is usually considered the more relevant result of the square root (the principle square root, as it's called):
(0 - 1/2) = (1 - 1/2)
...add t/2 to both sides, as we did above:
0 = 1

Because every number can be expressed as itself times the multiplicative identity (1), and 1 actually equals 0, then this means that every number is actually 0 in disguise.
For the numbers relevant here:
6 = 6 * 1 = 6 * 0 = 0
3 = 3 * 1 = 3 * 0 = 0
2 = 2 * 1 = 2 * 0 = 0
1 = 1 * 1 = 1 * 0 = 0
So all of these numbers equal each other and should be freely interchangeable; we can substitute any of these numbers in for any other one of these numbers.
So 2/6 = 1/3 is the same as stating 6/1 = 1/3, which is clearly false.

(I, for the record, know what's wrong with the above. Anyone else? It isn't division by zero -- that's amateur crap.)

[T3chno]

Haha. Kablizzy's still on the troll-mobile. Poor saps. :/

[t̷s͢uk̕a͡t͜ư]

Haha. Kablizzy's still on the troll-mobile. Poor saps. :/

If he thinks that the last few pages have had anything at all to do with his actions, that's his delusion.

[DemonzLunchBreak]

t = a + b
(a - b) t = (a - b) (a + b)
at - bt = a^2 - b^2
a^2 - at = b^2 - bt
a^2 - at + (t/2)^2 = b^2 - bt + (t/2)^2
(a - t/2)^2 = (b - t/2)^2
Two paths from here, since taking a square root gives two answers...
Taking the negative square root:
a - t/2 = -(b - t/2)
a - t/2 = t/2 - b
a + b = t
...this merely confirms our given, meaning that we didn't do anything wrong in the previous steps.
Taking the positive square root:
a - t/2 = b - t/2
a = b

Foiled incorrectly. (a - b)(a +b) = a^2 - 2ab + b^2. (a - b)(a + b) != a^2 + b^2.

[blue_tetris]

Might wanna recheck your math, Demonz.

(a - b)(a + b) = a^2 - 2ab + b^2. is wrong.

(a - b)(a + b) = a^2 - ab + ab - b^2 = a^2 - b^2.



What do I know, though. I'm a businessman.

Please tell me where you got -2ab, though. Because I see a * b and then I see -b * a. And none of that adds to -2ab.