Math Stuff

[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

[Morat]

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]

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]

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]

.99999 = 1

 

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

 

x = .9999999...

 

10x = 9.9999999...

 

- x = .99999999...

 

9x = 9

 

x = 1

[fafalone]

.111111... = 1/9

 

9*1/9 = 1

[JaKiri]

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]

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]

But how would 9*.111111111111111111111 add a zero?

[Epsilon]

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]

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]

One of my favorite equations is e^(pi*i) + 1 = 0

 

 

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

Just aman

[fafalone]

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]

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

[Guest taoist]

Why all women are evil ...

 

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

 

We know that ...

1. Women mean time and money.
   
2. Time is money.
   
3. 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]

Yea, I've seen the paper that has that written on it

 

[JaKiri]

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]

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".