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Old 01-12-2011, 04:24 PM   #1
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Check the math, please ...

16 - 36 = 25 - 45
Both result in -20, so they're equivalent, right?

(16 - 36) + 81/4 = (25 - 45) + 81/4
Since they're both equivalent, let's add 81/4 to each side

a^2 - 2*a*b + b^2 = (a - b)^2

or

4^2 - 2*4*(9/2) + (9/2)^2 = 5^2 - 2*5*(9/2) + (9/2)^2
(16 - 36 + 81/14 = 25 - 45 + 81/4)

4 - 9/2 = 5 - 9/2
Performed square root and subtraction from each side

4 = 5
Removed extraneous data (9/2) from each side

Therefore:

2+2 = 5

... right or wrong?
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Old 01-12-2011, 04:29 PM   #2
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wrong.
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Old 01-12-2011, 04:30 PM   #3
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Square root of negative one...
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Old 01-12-2011, 04:31 PM   #4
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both trains arrive at 4 pm.
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Old 01-12-2011, 04:32 PM   #5
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Think you forgot to run this past mrkris..
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Old 01-12-2011, 04:33 PM   #6
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Old 01-12-2011, 04:42 PM   #7
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a^2 - 2*a*b + b^2 = (a - b)^2

That would be a brand new kind of math than I never learned.
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Old 01-12-2011, 04:44 PM   #8
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a^2 - 2*a*b + b^2 = (a - b)^2

That would be a brand new kind of math than I never learned.
where did you go to school?
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Old 01-12-2011, 04:48 PM   #9
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(4 - 9/2)^2=(5 - 9/2)^2
|4 - 9/2|=|5 - 9/2|
4 - 9/2=+\-(5 - 9/2)
4 - 9/2 = -5 + 9/2
4 + 5 = 9/2 + 9/2
9 = 9
2 + 2 = 4 !!!
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Old 01-12-2011, 04:49 PM   #10
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a^2 - 2*a*b + b^2 = (a - b)^2

That would be a brand new kind of math than I never learned.
i could be wrong, but from my trig days, i believe this is the pythagorean formula.
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Old 01-12-2011, 04:51 PM   #11
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where did you go to school?
Some dumb ass college that told me to solve the problem by isolating "a" and "b" and
defining "a" in terms of "b" or "b" in terms of "a" and then plugging in either a value for "a" or "b" and then solving the equation for the other variable.


Yeah, wish I could get my money back for that course.

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Old 01-12-2011, 04:54 PM   #12
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Some dumb ass college that told me to solve the problem by isolating "a" and "b" and
defining "a" in terms of "b" or "b" in terms of "a" and then plugging in either a value for "a" or "b" and then solving the equation for the other variable.


Yeah, wish I could get my money back for that course.

As far as I remember its about 6th grade material, not college stuff
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Old 01-12-2011, 04:55 PM   #13
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a^2 - 2*a*b + b^2 = (a - b)^2

That would be a brand new kind of math than I never learned.
(a - b)^2 = (a - b)*(a - b) = a*a - a*b - b*a + b*b = a^2 - a*b - a*b + b^2 = a^2 - 2a*b + b^2

a^2 - 2*a*b + b^2 = (a - b)^2
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Old 01-12-2011, 04:56 PM   #14
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are you all motherfuckers on crack or what
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Old 01-12-2011, 04:56 PM   #15
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2 + 2 = 4
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Old 01-12-2011, 04:59 PM   #16
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i could be wrong, but from my trig days, i believe this is the pythagorean formula.
Oh yeah,

Like a^2 + b^2 = c^2

It's like a^2 is one side of the triangle and b^2 is another side and c^2 is the hypotenuse.

what happens if you have value 7=c^2 and try to solve for c?
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Old 01-12-2011, 05:10 PM   #17
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2+2=5 or 2*2=5 - this equates to only when you go to the bank to take credit! ;)
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Old 01-12-2011, 05:15 PM   #18
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Hehehe, this is an old troll, but still a pretty fun one. Usually it ends with "WHERE IS YOUR GOD NOW?!"
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Old 01-12-2011, 05:35 PM   #19
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(a - b)^2 = (a - b)*(a - b) = a*a - a*b - b*a + b*b = a^2 - a*b - a*b + b^2 = a^2 - 2a*b + b^2

a^2 - 2*a*b + b^2 = (a - b)^2
I got this when I tried that :

(a - b)^2 = (a - b)*(a - b) = a(a - b) - b(a - b) = (a^2 - ab) - (ab - b^2) = a^2 - 2ab - b^2
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Old 01-12-2011, 05:47 PM   #20
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I got this when I tried that :

(a - b)^2 = (a - b)*(a - b) = a(a - b) - b(a - b) = (a^2 - ab) - (ab - b^2) = a^2 - 2ab - b^2
please stop ..
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Old 01-12-2011, 06:04 PM   #21
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math doesn't suck...
but does she (suck)?
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Old 01-12-2011, 07:05 PM   #22
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please stop ..
No, I must keep going until the truth is exposed.
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Old 01-12-2011, 09:08 PM   #23
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Oldie but a goodie.

Much simpler version with a=2 b=1 and easier to follow.

I didn't check it, but I'm 99% sure you introduced an irrelevant root.

No use making a joke at this point as Americans don't Root!


CLUE

LET X = -2
X^2 = 4
X = SQRT(4)
X = 2

2 = -2
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Old 01-13-2011, 02:45 AM   #24
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Old 01-13-2011, 03:06 AM   #25
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So you've got

let
a^2 - 2ab + b^2 = c^2 -2cd + d^2

therefore
(a - b)^2 = (c - d)^2

therefore
(a - b) = (c - d)

let a = 1, b = 2, c =4, d = 3

(1 - 2)^2 = (4 - 3)^2

1 = 1

Even though (1 - 2) =/= (4 - 3)

so X^2 = Y^2
-/-> X = Y
(equal squares does not imply equal values)

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Old 01-13-2011, 03:12 AM   #26
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Heres one problem thats elementary and uses no tricks.
1/3 = 0.(3) (0.(3) means 0 followed by an infinite number of 3s 0.333333333333...)

1/3 + 1/3 + 1/3 = 3/3 = 1

0.(3) + 0.(3) + 0.(3) = 0.(9)

results:

1 = 0.(9) !
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Old 01-13-2011, 03:36 AM   #27
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There is no possible number between 0.999999... and 1

so they must be equal.
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Old 01-13-2011, 03:45 AM   #28
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There is no possible number between 0.999999... and 1

so they must be equal.
0.(9) = 9/10
you're saying that
9/10 = 1 ? which equivs 9 = 10
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Old 01-13-2011, 04:02 AM   #29
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0.9... = 9/9
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Old 01-13-2011, 04:04 AM   #30
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this is like playing some residet evil puzzle its so hard!!
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Old 01-13-2011, 04:10 AM   #31
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wrong )
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Old 01-13-2011, 04:16 AM   #32
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0.9... = 9/9
math is not something you make up on the way. If thats what you've been thought you should slap your math teacher.
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Old 01-13-2011, 04:26 AM   #33
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0.(9) = 9/10
you're saying that
9/10 = 1 ? Which equivs 9 = 10
0.(9) != 9/10
0.9 = 9/10
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Old 01-13-2011, 05:32 AM   #34
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1/9 = 0.11111...
2/9 = 0.22222...
3/9 = 0.33333...
4/9 = 0.44444...
5/9 = 0.55555...
6/9 = 0.66666...
7/9 = 0.77777...
8/9 = 0.88888...
9/9 = 0.99999...




Seems like a useful skill!
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Old 01-13-2011, 05:34 AM   #35
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i was always bad at math.
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Old 01-13-2011, 05:39 AM   #36
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1/9 = 0.11111...
2/9 = 0.22222...
3/9 = 0.33333...
4/9 = 0.44444...
5/9 = 0.55555...
6/9 = 0.66666...
7/9 = 0.77777...
8/9 = 0.88888...
9/9 = 0.99999...
While the first 8 match, the 9th is made so your logic works because 9/9 =1 not 0.(9) because you say so.

Quote:


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fuck off mister lunatic. Don't you need to count your rent money
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Old 01-13-2011, 05:40 AM   #37
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0.(9) != 9/10
0.9 = 9/10
my bad on that one
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Old 01-13-2011, 05:46 AM   #38
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Huh? I haven't been corrected in 10,000 posts on sci.math in 10 years.

This is kindy stuff. Stop twisting my posts.
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Old 01-13-2011, 05:48 AM   #39
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Huh? I haven't been corrected in 10,000 posts on sci.math in 10 years.

This is kindy stuff. Stop twisting my posts.
Obviously, also nobody corrected you in your paranormal bullshit in hundreds of posts
mister Truman
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Old 01-13-2011, 05:54 AM   #40
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So tell me a number BETWEEN 0.9.. recurring and 1.0 Smartass!
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Old 01-13-2011, 06:06 AM   #41
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Old 01-13-2011, 06:13 AM   #42
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so tell me a number between 0.9.. Recurring and 1.0 smartass!
1 - 0.(9)

In all seriousness I've read this started a lot of debates so I'm going to let it go https://secure.wikimedia.org/wikipedia/en/wiki/0.999...
To me it seems that its just made to fit in so things work, but hey
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Last edited by darksoul; 01-13-2011 at 06:16 AM..
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Old 01-13-2011, 06:44 AM   #43
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yeh right SMART ASS!

ADAM WINS AGAIN! 100% RECORD
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Old 01-13-2011, 07:09 AM   #44
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yeh right SMART ASS!

ADAM WINS AGAIN! 100% RECORD
you know the saying.
even if you win, you're still retarted.
I pity you
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Old 01-13-2011, 07:54 AM   #45
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being reTarted doesn't sound so bad! Looking forward to it in fact!
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