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Seagull
11-04-2003, 12:34 PM
John likes 400 but not 300; he likes 100 but not 99; he likes 2500 but not 2400. Which does he like?
900
1000
1100
1200

zenbooty
11-04-2003, 12:41 PM
900. Johnny's into the squares.

Seagull
11-04-2003, 01:34 PM
I knew it was something easy, I got hung up looking for a common denominator.

kei2
11-04-2003, 01:35 PM
Man, Zen's always got these riddle/puzzle answers in a hurry. :bow:

zenbooty
11-04-2003, 02:56 PM
Originally posted by kei2
Man, Zen's always got these riddle/puzzle answers in a hurry. :bow: :D Me likes games and puzzles! Here's a toughy I been tryin to figure out:

Readers of Prime Obsession will know that, contrary to what your schoolteachers told you, the number "minus one" does so have a square root, the friendly little number i. In fact, not only is the square of i equal to -1, so, by the rule of signs, is the square of -i.

If you are still with me, see if you can find the flaw in the following argument. Out of consideration for Aaron the Webbie Master, I am going to use "Sqr(x)" to indicate the square root of x and an asterisk to indicate multiplication.

Start from this obvious truth: Sqr(x-y) = I * Sqr(y — x).

Since this is plainly the case for any numbers x and y, put x = a and y = b. Then

Sqr(a — b) = i * Sqr(b — a).

On the other hand, since that obvious truth I started with is the case for any numbers x and y, I could equally well put x = b and y = a, to get another, equally true statement:

Sqr(b — a) = i * Sqr(a — b).

Now, if P = Q and R = S, then obviously P * R = Q * S. So:

Sqr(a — b) * Sqr(b — a) = i^2 * Sqr(b — a) * Sqr(a — b).

Since i^2 = -1 and the other components of each side are identical, it follows that 1 = -1. It easily follows that every minus sign can be replaced by a plus, all debits are credits, all liabilities are assets, and the pope is Jewish. (Proof of the latter: Since 1 = -1, adding 1 to each side gives 2 = 0. Halving both sides, 1 = 0. Adding 1 to both sides, 2 = 1. Now, the pope and Jackie Mason — who is Jewish — are two people, therefore they are one person, therefore the pope is Jewish.)

Where is the logical flaw?

11-04-2003, 03:17 PM
is it really that hard? i think i know the answer .... i'll use >> for your quotes:

>> Start from this obvious truth: Sqr(x-y) = I * Sqr(y — x).
>>

TRUE.

>> Since this is plainly the case for any numbers x and y,
>> put x = a and y = b. Then
>>
>> Sqr(a — b) = i * Sqr(b — a).
>>

substitution, ok...

>> On the other hand, since that obvious truth
>> I started with is the case for any numbers x and y,
>> I could equally well put x = b and y = a, to get another,
>> equally true statement:
>>
>> Sqr(b — a) = i * Sqr(a — b).
>>

this is where a flaw occurs. yes, you can substitute ANYTHING (anything i THINK) for x,y.

>> Now, if P = Q and R = S, then obviously P * R = Q * S. So:
>>
>> Sqr(a — b) * Sqr(b — a) = i^2 * Sqr(b — a) * Sqr(a — b).

yes, you can multiply like that. BUT you need to realize this...

part 1 you have x=a, y=b. that's ok. second part you substitute x=b, y=a.

if b=b and a=a from both parts, then a=b. why?
p1: sqr(a-b) = i*sqr(b-a), x=a, y=b
p2: sqr(b-a) = i*sqr(a-b), x=b, y=a
so: a=b
(i'm too tired to pull up some REAL proofs)

thus, you can say
Sqr(a — b) * Sqr(b — a) = i^2 * Sqr(b — a) * Sqr(a — b) where a=b (proved above) cause 0 = 0 --> sqr(a-a) = 0

or, if you say they don't have to be the same, then using same letter variable you need to sub-number them so in part 1 x=a1, y=b1, and part 2 has x=b2, y=a2
where
sqr(a1-b1)*sqr(b2-a2) = i^2 *sqr(b1-a1)*sqr(a2-b2)

ski
11-04-2003, 03:52 PM
Hey Seagull.... OT, but where in Central VA are you? I don't see many Virginians on here... I'm in Lynchburg.

And zen -- wow on that one you've been trying to figure out :eek: :)

zenbooty
11-04-2003, 04:04 PM
Sorry Dr. Jam, but I don't think you're right. The fact that you can switch a and b as values for x and y doesn't imply a=b. The function is a truism for all values of x and y, including both permutations with a and b. So the equation is true for x=a and y=b, but is also true for x=b and y=a, regardless of the value of a and b. And the two resulting equations don't have anything to do with one another as far as proving a=b.

To better explain what I'm trying to say, lets use a simpler example than the equation in the puzzle. Sorta like this rule:

x + y = y + x

This is true for all x and y, right? So, we could substitute x=a and y=b and it would be true, or substitute x=b and y=a, and it would still be true, without implying that a=b, right? If you don't believe me, look at it with x=4 and y=3, or any other real values you want.

No I think the error has to have something to do with the initial "obvious" truism upon which the whole puzzle is base. Namely, that you can extrapolate i from a squareroot by dividing the square-rooted value by -1. On its face it looks right, because that's what you would do with positive real values. But there's gotta be something wrong with doing it with a negative. I just can't figure out, or remember, exactly what.

RoniMan
11-04-2003, 04:10 PM
Originally posted by zenbooty

Now, if P = Q and R = S, then obviously P * R = Q * S. So:

Sqr(a — b) * Sqr(b — a) = i^2 * Sqr(b — a) * Sqr(a — b).

Since i^2 = -1 and the other components of each side are identical, it follows that 1 = -1. It easily follows that every minus sign can be replaced by a plus, all debits are credits, all liabilities are assets, and the pope is Jewish. (Proof of the latter: Since 1 = -1, adding 1 to each side gives 2 = 0. Halving both sides, 1 = 0. Adding 1 to both sides, 2 = 1. Now, the pope and Jackie Mason — who is Jewish — are two people, therefore they are one person, therefore the pope is Jewish.)

Where is the logical flaw?

i'll give it a shot!

i think the flaw lies in the the fact that Sqr(x) results in a + or - answer. therefore, P*R could be either + or - as well as Q*S.

oh yeah, just in case that explanation wasn't clear.. sqr (25) = -5 OR +5, and sqr(-25) = -5i, or +5i

zenbooty
11-04-2003, 04:25 PM
Originally posted by RoniMan
i'll give it a shot!

i think the flaw lies in the the fact that Sqr(x) results in a + or - answer. therefore, P*R could be either + or - as well as Q*S.

oh yeah, just in case that explanation wasn't clear.. sqr (25) = -5 OR +5, and sqr(-25) = -5i, or +5i Yeah that's what I've been thinking. But now how to take that and correctly apply it to the problem to show how its logic fails. Making the connection is what I'm trying to figure out.

RoniMan
11-04-2003, 04:32 PM
well, try this (from this point +- is the sign for plus/minus
Sqr(a — b) = +-i * Sqr(b — a)
and Sqr(b — a) = +-i * Sqr(a — b)

therefore, your new statement has to be P=+-Q, and R=+-S

does that help?

Kenas
11-04-2003, 04:38 PM
Originally posted by zenbooty

Since this is plainly the case for any numbers x and y, put x = a and y = b. Then
...
On the other hand, since that obvious truth I started with is the case for any numbers x and y, I could equally well put x = b and y = a

These two statement make x = a = b = y so when you come down to the equation
Sqr(a — b) * Sqr(b — a) = i^2 * Sqr(b — a) * Sqr(a — b). The same as
Sqr(a - a) * Sqr(a - a) = i^2 * Sqr(a - a) * Sqr(a - a)
OR
0 = 0

RoniMan
11-04-2003, 04:40 PM
i think that's what djradam was trying to say, try reading zen's explanation for that one. it's base upon the commutative property of addition.

le_stick
11-04-2003, 06:30 PM
You guys make my head dizzy.....I have to take a break now.....its all your fault....heheee

bachviet
11-04-2003, 06:46 PM
Damn math...

InfiniteNothing
11-04-2003, 06:50 PM
Originally posted by Kenas

These two statement make x = a = b = y so

No they don't

The problem is the premis. It is only true when x<y. So you know that one of either the second or the third statement is false. (since in one of them x < y)

another way to see it is in the i^2 at the end. I=sqrt(-1) so sqrt(-1)*sqrt(-1). Since sqrt (a)* sqrt(b) = sqrt(a*b), sqrt(-1)* sqrt(-1)= sqrt(-1*-1)=sqrt(1)=1 thus i^2 =1 making the statement M=M

Mrs. Happy Squirrel
11-04-2003, 07:27 PM
I agree with everybody else.. all those letters and numbers.. and all of these letters and numbers.. how am i even typing? the universe has collapsed... iiiiiiiieeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

and it's all math's fault... binarys and all.... the irony, oh the sweet irony........

Seagull
11-04-2003, 10:46 PM
Originally posted by skiAtomic
Hey Seagull.... OT, but where in Central VA are you? I don't see many Virginians on here... I'm in Lynchburg.

I'm up in the Charlottesville area, my wife is from the Lynchburg area.

11-04-2003, 11:16 PM
sorry infinit, i^2 = -1 . on a side note: sqrt(a)*sqrt(b) = sqrt(a*b) occurs when a,b are non-negative. i think we need a,b to be ANY number. (ref: http://mathforum.org/library/drmath/view/53813.html).

i may have overlooked something, but I still think that a=b in that case. anyway, i'll try again.

let X = a - b, Y = b - a
X != x, Y != y (look at cases here)

part one (given)
Sqr(a — b) = i * Sqr(b — a) ===>>>>
Sqr(X) = i * Sqr(Y) ::: 1
check.

part 1a
Sqr(X) = i * Sqr(Y)
Sqr(X) / i = i * Sqr(Y) / i
Sqr(X) / i = Sqr(Y)
** 1 / i = -i
-i * Sqr(X) = Sqr(Y) ::: 1a
-i * Sqr(a-b) = Sqr(b-a) ::: 1a

part 2 you have

Sqr(b — a) = i * Sqr(a — b) ===>>>>
Sqr(Y) = i * Sqr(X)
Sqr(b-a) = i * Sqr(a-b)

this does not follow what we have concluded in 1a. there is the flaw.
when you plug it all in again using Sqr(b-a)= -i * Sqr(a-b), you get the right answer, 1=1

THAT should work for now at least....

InfiniteNothing
11-05-2003, 12:05 AM
sorry infinit,

I'm right. Ignore all the I^2 = 1 crap. The premise is not true! Just plug in an y<x and you'll see.

11-05-2003, 12:26 AM
Originally posted by InfiniteNothing

I'm right. Ignore all the I^2 = 1 crap. The premise is not true! Just plug in an y<x and you'll see.

my bad, the sqrt(a)*sqrt(b)=sqrt(a*b) when BOTH are not negative. there are exceptions i guess... ex: a=-4, b=-9 -> -6 != 6

i guess from what you said above, the second statement is false, and thus the third statement is false.

InfiniteNothing
11-05-2003, 12:36 AM

i may have overlooked something, but I still think that a=b in that case. anyway, i'll try again.

let X = a - b, Y = b - a
X != x, Y != y (look at cases here)

part one (given)
Sqr(a — b) = i * Sqr(b — a) ===>>>>
Sqr(X) = i * Sqr(Y) ::: 1
check.

part 1a
Sqr(X) = i * Sqr(Y)
Sqr(X) / i = i * Sqr(Y) / i
Sqr(X) / i = Sqr(Y)
** 1 / i = -i
-i * Sqr(X) = Sqr(Y) ::: 1a
-i * Sqr(a-b) = Sqr(b-a) ::: 1a

part 2 you have

Sqr(b — a) = i * Sqr(a — b) ===>>>>
Sqr(Y) = i * Sqr(X)
Sqr(b-a) = i * Sqr(a-b)

this does not follow what we have concluded in 1a. there is the flaw.
when you plug it all in again using Sqr(b-a)= -i * Sqr(a-b), you get the right answer, 1=1

THAT should work for now at least....

First off a need not = b, see zen post or take it this way

you agree that

Sqrt(x — y) = i * Sqrt(x — y)
and just messing around with letters
Sqrt(s-t) = i * Sqrt (t - s)

Now lets say x = a y=b and s=b and t=a
then you can say that x=t but not that a =b
now substituting
Sqr(b — a) = i * Sqr(a — b).
Sqr(a — b) = i * Sqr(b — a).
You've already agreed this is true. It's just substitution.
And the rest of the proof follows normally.

You found an error but the author did too. That doesn't disprove the statement (that is, it doesn't disprove substitution), it just implies their's a mistake somewhere in there but we already knew that since 1 =! -1. Infact if you switch the right side of equation 1a with the left side and compare it to the last equation you REPROVE the author's fallacy.

my bad, the sqrt(a)*sqrt(b)=sqrt(a*b) when BOTH are not negative. there are exceptions i guess... ex: a=-4, b=-9 -> -6 != 6

i guess from what you said above, the second statement is false, and thus the third statement is false.

RIGHT!!!!! That's where the logic error comes from. That formula was used to find the premise and hence the logic error is before the riddle begins. It's ironic because we are taught that [equation] in school. But if you assume it's correct (I know it's not) it's neat to see how I^2 ends up = 1 making the statment M= I^2 M true

cheapchinese
11-05-2003, 02:00 AM

Jcranmer
11-05-2003, 07:58 AM
sorry infinit, i^2 = -1 . on a side note: sqrt(a)*sqrt(b) = sqrt(a*b) occurs when a,b are non-negative. i think we need a,b to be ANY number. (ref: http://mathforum.org/library/drmath/view/53813.html).

i may have overlooked something, but I still think that a=b in that case. anyway, i'll try again.

let X = a - b, Y = b - a
X != x, Y != y (look at cases here)

part one (given)
Sqr(a — b) = i * Sqr(b — a) ===>>>>
Sqr(X) = i * Sqr(Y) ::: 1
check.

part 1a
Sqr(X) = i * Sqr(Y)
Sqr(X) / i = i * Sqr(Y) / i
Sqr(X) / i = Sqr(Y)
** 1 / i = -i
-i * Sqr(X) = Sqr(Y) ::: 1a
-i * Sqr(a-b) = Sqr(b-a) ::: 1a

part 2 you have

Sqr(b — a) = i * Sqr(a — b) ===>>>>
Sqr(Y) = i * Sqr(X)
Sqr(b-a) = i * Sqr(a-b)

this does not follow what we have concluded in 1a. there is the flaw.
when you plug it all in again using Sqr(b-a)= -i * Sqr(a-b), you get the right answer, 1=1

THAT should work for now at least....

Please show me one real world use for the above greek please, other then physics or rocket science. Or a math teacher. :D

zenbooty
11-05-2003, 08:29 AM
Originally posted by Jcranmer

Please show me one real world use for the above greek please, other then physics or rocket science. Or a math teacher. :D If the real world is what interests you, what are you doing on G|A?

11-05-2003, 04:20 PM
oh my, my bad infinite, i skipped over some posts that would have been helpful: roni's/yours. duh, been tired working so much! i think to relate the
sqrt = +/- value property, it can be put together with what i said earlier.

the error would occur in the second statement.

If
(1) Sqr(b - a) = i * Sqr(a - b), then:

(2) Sqr(a - b) != i * Sqr(b - a).
The correct statement would be:

(3) Sqr(a - b) = -i * Sqr(b - a).

Multiply both sides of (3) by i and that would give
you the original statement (1). The reason the second
statement is false is because the square root of
something has both positive and negative values, which i overlooked and all that.

molecularfire
11-05-2003, 06:03 PM
Man... there is a lot of static in that question. The simple answer is that you can't divide by zero.

revil
11-05-2003, 06:14 PM
um... by just briefly looking at it, here is the logic error that makes sense in my head:

Sqr(a — b) = i * Sqr(b — a).

On the other hand, since that obvious truth I started with is the case for any numbers x and y, I could equally well put x = b and y = a, to get another, equally true statement:

Sqr(b — a) = i * Sqr(a — b).

Now, if P = Q and R = S, then obviously P * R = Q * S. So:

Sqr(a — b) * Sqr(b — a) = i^2 * Sqr(b — a) * Sqr(a — b).

This thing is getting at that Sqr(a — b) == Sqr(b — a) by fooling around with the letters. this can't be. it's either one way or the other way. it can't be both:
ie: x=a and y=b or x=b and y=a, not x=a=b and y=b=a

though that's just from glancing at it.

zenbooty
11-05-2003, 08:03 PM

InfiniteNothing
11-05-2003, 11:25 PM
Originally posted by molecularfire
Man... there is a lot of static in that question. The simple answer is that you can't divide by zero.

What?

Dj:
No, the error is in the premise..damn, I sound like a parrot.

In your example the third statment is true and the second statment is only false when a > b. If a < b statment 2 is correct but statement 1 is incorrect. But that's besides the point. If you admit the premis s(x-y) = i s(y-x), you must admit both statment's 1 and 2 are correct because they are just diffrent forms of the premise.

revil
11-06-2003, 12:22 AM
.

11-06-2003, 12:45 AM
the premise should be

s(x-y)= +/- i*s(y-x)

right?

InfiniteNothing
11-06-2003, 01:45 AM
That is a correct premise. I'd say

{ s(x - y) = i s(y-x) when x < or = y
s(x - y) = -i s(y-x) when x > y }

If you do the proof like that you end up with M=M but your statment is accurate too. Though I think you'll get M= +/-M when you do the proof and that may be wrong, I don't know. Is it correct to say 1 + 1 = +/- 2

Sir_Froggy
11-06-2003, 01:52 AM
Originally posted by Yossarian
ok, all that math makes my head hurt...math blows, thats why i'm a history buff

me too, but not a major history buff

oh on that note, I was watching who wants to be a millionaire before school the other day, and the second to last question was a little hard, but I still got it right, but the last question was so easy, oh this guy got the million, anyway do they always put an easy question last?

11-06-2003, 03:03 PM
Originally posted by InfiniteNothing
That is a correct premise. I'd say

{ s(x - y) = i s(y-x) when x < or = y
s(x - y) = -i s(y-x) when x > y }

If you do the proof like that you end up with M=M but your statment is accurate too. Though I think you'll get M= +/-M when you do the proof and that may be wrong, I don't know. Is it correct to say 1 + 1 = +/- 2

true, i don't know the exact notation... cause 1+1 = +/- 2 i think would mean 1+1 = 2 or 1+1=-2

but then i wonder how you would fit what you just said into the equation, when you multiply them together...

RoniMan
11-06-2003, 03:13 PM

true, i don't know the exact notation... cause 1+1 = +/- 2 i think would mean 1+1 = 2 or 1+1=-2

but then i wonder how you would fit what you just said into the equation, when you multiply them together...

umm..i thought we solved this already...sqr 1 = +-1

therefore 1 + 1 = 2, -1 + -1 = -2.

welfareloser
11-06-2003, 03:19 PM
this thread no longer amuses us. desist.

zenbooty
11-06-2003, 04:06 PM
Originally posted by welfareloser
this thread no longer amuses us. desist. Spoilsport. :pfft: :pfft: :pfft:

:P

InfiniteNothing
11-06-2003, 04:27 PM
Originally posted by RoniMan

umm..i thought we solved this already...sqr 1 = +-1

therefore 1 + 1 = 2, -1 + -1 = -2.

What? First off sr(1) is not +- 1.

No I was saying is "1+1 = +-2" a correct statment?