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blike
July 27th, 2002, 2:49 PM
Just for fun.

Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0

--

1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c

--

Proof:
1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1^ = -1

Also one can disprove the axiom that things equal to the same thing
are equal to each other.

1 = sqrt(1)
-1 = sqrt(1)
therefore 1 = -1

Morat
July 27th, 2002, 3:41 PM
Originally posted by blike
Just for fun.

Proof: Suppose that a=b. Then
a = b
a^2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
a + b = b
a = 0

--

1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c

--

Proof:
1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1^ = -1

Also one can disprove the axiom that things equal to the same thing
are equal to each other.

1 = sqrt(1)
-1 = sqrt(1)
therefore 1 = -1
1) Divide by zero error
2) $1 != (10c)^2, $1 = (10^2)c
3) I'm not sure, i think the axiom applies to defined terms only, not functions with multiple values.

aman
July 27th, 2002, 4:13 PM
An anology. I have a pound of peanuts = 1lb.
Joe owes me a pound of peanuts =1lb.
1lb I have does not equal 1lb I'm owed.
That's the first thing that pops in my head but I ain't done yet.
This is fun:bravo:
Just aman

fafalone
July 27th, 2002, 4:39 PM
sqrt(a*b) does not equal sqrt(a)*sqrt(b)

(a*b)^(1/2) cannot be expanded (look at sqrt as this power, not a function)

blike
July 27th, 2002, 4:49 PM
.99999 = 1

------------------

x = .9999999...

10x = 9.9999999...

- x = .99999999...

9x = 9

x = 1

fafalone
July 27th, 2002, 5:21 PM
.111111... = 1/9

9*1/9 = 1

JaKiri
July 27th, 2002, 5:23 PM
Originally posted by blike
.99999 = 1

------------------

x = .9999999...

10x = 9.9999999...

- x = .99999999...

9x = 9

x = 1

That's right though, 0.9 rec. is equiv to 1.

(this is in relevence to

Giles says:
whats the problem with blike's last proof?

)

aman
July 27th, 2002, 6:52 PM
The .9999... no matter how far it extends has to move up one decimal place. The ...... is confusing until you realize moving it one digit past the decimial point is also adding a zero in the last place at the infinite end. Then you are subtracting

.9999-----0 from

.9999-----9

That last 9 tips the value to 1
Just aman

fafalone
July 27th, 2002, 7:32 PM
But how would 9*.111111111111111111111 add a zero?

Epsilon
July 27th, 2002, 9:04 PM
sqrt(ab) does equal sqrt(a) * sqrt(b), equivalently (ab)^(1/2) = a^(1/2)*b^(1/2)

Of course when a,b < 0 you run into problems.

sqrt(-25) = sqrt(25*-1) = sqrt(25)*sqrt(-1) = 5i, where i is sqrt(-1)

But when one factor is negative and one is positive you're fine.

For instance: sqrt(-5) * sqrt(-5) = -5, because by definition (sqrt(x))^2 = x. However if you group (for lack of a better term) the power 1/2 over the factors which are negative you run into problems....

sqrt(-5) * sqrt(-5) as sqrt(-5*-5) as sqrt(25) = 5.

Anyway, in short, the rules of radicals for real numbers are not explicitly the same for complex.

If that didn't make sense, I'll let someone else try to explain. For the most part, I think it is a matter of definitions of radicals which give conflicts with negative numbers.

---------------

0.99999... does not terminate. Do not look at it as a finite decimal.

0.99999... is a converging geometric series. It's in the form of S_n = a_1/(1-r), where S_n is the sum of the nth terms in the series, and a_1 is the first term and r is the ratio between each terms a_(n+1)/a_n

0.99999... converges (approaches a limit) at 1, hence if the number of terms in this particular series goes into infinity (indicated by the "..." notation) the series can never be greater than 1 so we say that 0.999... = 1


Anyway, if I missed something again, let me know.

fafalone
July 27th, 2002, 9:32 PM
Can't say it's equal unless you constrain it to numbers larger than 0. And it my point was it's not true with arguments less than 0.

i is rather difficult to explain in detail, so if you don't understand it let me know before I get too annoyed trying to find a simple answer.
One of my favorite equations is e^(pi*i) + 1 = 0, incorporating the 5 most important values in math :)

aman
September 23rd, 2002, 2:51 PM
One of my favorite equations is e^(pi*i) + 1 = 0


Please explain what "i" is in this equation. Thanks
Just aman

fafalone
September 23rd, 2002, 4:51 PM
i is the square root of -1, an "imaginary" number.

sqrt(4) = 2
sqrt(-4) = 2i

i^2=-1

e^i = cis(1) (cis is the function cos(x)+i*sin(x))

another interesting property is i^i = e^(-:pi:/2)

ln(-x) = ln(x) + :pi:i

aman
September 23rd, 2002, 11:07 PM
Thanks Faf. It's been twenty years since I had my head in a school book so I appreciate your patience
Thanks for the equations. Math can go where no one has gone before and even when you sitting down or trying to get to sleep.
It's just that some of it makes sense mathematicaly but not logically or is at least difficult.
Just aman

taoist
May 22nd, 2003, 8:13 PM
Why all women are evil ...

(given me by a FEMALE colleague, so don't even say it)

We know that ...
Women mean time and money.
Time is money.
Money is the root of all evil.But that means women ...

= time * money (by 1)
= money ^ 2 (by 2)
= evil (by 3)

(of course, she was hoping it wouldn't check out)

blike
May 22nd, 2003, 9:16 PM
Yea, I've seen the paper that has that written on it :-p

http://www.math.yorku.ca/infinity/fun/evilgirls.jpg

JaKiri
May 22nd, 2003, 9:17 PM
Originally posted by blike
Proof:
1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1^ = -1

Also one can disprove the axiom that things equal to the same thing
are equal to each other.

1 = sqrt(1)
-1 = sqrt(1)
therefore 1 = -1

As we're bumping old threads...

You need moduli for dem.

|1| does indeed = |-1|.

It's basically the same as

log 1 = log (-1)^2 = 2 log (-1)

0 = 2 log (-1)

log (-1) = 0

log (-1) = log (1)

1 = -1.

Again, missing the moduli.

Tom Mattson
May 28th, 2003, 6:24 PM
I know you guys are already way past this first one, but I have to point something out that everyone missed.

Originally posted by blike
1$ = 100c
= (10c)^2
= (0.1$)^2
= 0.01$
= 1c

--


There is a glaring mistake above, and it has to do with units. When you square (0.1$), you get (0.01$2), or 0.01 "square dollars".