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Author
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
windwakr wrote:
I WANT ANOTHER STAR lol
Hehe, yes I understand, they are very valuable and greatly sought after. BUT, you still have to find the solution. The line above you state as dividing by (pi-3) is actually expanding (by multiplying out the factors) the equalities.
12 Aug 2009, 05:42
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
Ok, I see(NOT)...lol, math is not my strong point.

EDIT:DAMMIT...Forget that last part i said...which I edited out, so hopefully noone saw it...somehow I was messing with the formula and forgot to change it back before looking into it again....

Hmmm, I must be dumb, it looks fine now that I've had more time to examine it. The only thing I see is that you ASSUME pi is equal to 3 at the beginning, else it fails later.

OK, at risk of brain explosion, I'll leave this alone.

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12 Aug 2009, 05:44
sleepsleep

Joined: 05 Oct 2006
Posts: 8904
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sleepsleep
i reexamine the oddities,
i think i found something weird here.

Quote:

2pix - 6x = pi^2 - 9
9 - 6x = pi^2 - 2pix
9 - 6x + x^2 = pi^2 - 2pix + x^2

i don't like pi, so i change it to y for better reading.

2xy - 6x = y^2 - 9
2xy - 6x + 9 = y^2 - 9 + 9
2xy - 2xy - 6x + 9 = y^2 - 2xy
-6x + 9 = y^2 - 2xy
-6x + 9 + x^2 = y^2 - 2xy + x^2
(x-3)(x-3) = (y^2 - x)

we expand (x-3)^2
x^2 - 3x - 3x + 9 or (x^2 - 6x + 9)

if in your weird example page 5.
u use (3-x)^2 which will become
9 - 3x - 3x + x^2 or (9 - 6x + x^2)

now is, (x^2 - 6x + 9) is (9 - 6x + x^2), but based on their arrangement,
we would either get (x-3)^2 or (3-x)^2

2xy - 6x = y^2 - 9
then u did some weird things into
9-6x = y^2 - 2xy <-------------- invalid.
it should be
-6x + 9 = y^2 - 2xy
12 Aug 2009, 09:40
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
sleepsleep wrote:
9-6x = y^2 - 2xy <-------------- invalid.
it should be
-6x + 9 = y^2 - 2xy
In a mathematical sense they are the same thing. Addition is commutative.
12 Aug 2009, 09:44
sleepsleep

Joined: 05 Oct 2006
Posts: 8904
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sleepsleep
then is, (x-3)^2 equal to (3-x)^2 ??
because,

using ur page 5
3-x = y -x
3 = y

if
x - 3 = y - x
x + x - 3 = y
2x - 3 = y (which is correct) because in ur page 5 line 2
2x = y + 3

so, it is errror to make it as (3-x)^2 = (y-x)^2 in ur example,
it should be (x-3)^2 = (y-x)^2
12 Aug 2009, 09:55
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
(3-x)^2 == (x-3)^2

Try it with any number you choose for x.
12 Aug 2009, 10:08
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
So far Borsuc has come the closest to the answer but just couldn't seem to make the last step.

And sleepsleep looks to be getting closer. Don't give up now, you can do it.
12 Aug 2009, 11:12
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
AHA, I have figured it out, I'm nearly positive...lemme just look over it a little longer.

EDIT:
3 - x= pi - x
Should be changed to:
x-3= pi - x

Which comes out to:
2x = pi+3

Code:
```           x = (pi + 3) / 2

2x = pi + 3

2x(pi - 3) = (pi + 3)(pi - 3)

2pix - 6x = pi^2 - 9

9 - 6x = pi^2 - 2pix

9 - 6x + x^2 = pi^2 - 2pix + x^2

(x - 3)^2 = (pi - x)^2   <----from here down is changed

x - 3 = pi - x

2x=pi+3

pi = 2x-3
```

(3-x)^2 and (x-3)^2 become whole different things on the line where the sqrt is taken.

EDIT:Just saw sleepsleep's post. Looks like you're trying to mess with him? Ya, (x-3)^2 and (3-x)^2 are equal by themselves, but when you take the square root of them they become different. x-3 is not equal to 3-x even with absolute, because of the ordering of the equation.

Right?

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Last edited by windwakr on 12 Aug 2009, 20:43; edited 7 times in total
12 Aug 2009, 18:10
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
Bump in case you didn't see my edit.
12 Aug 2009, 19:30
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution

Everybody is so close now. Should I be giving clues, or do you all enjoy the challenge more without external aids?
12 Aug 2009, 20:45
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
Still a problem?? ARRGGHHHH.......
lol, I thought I had it.

EDIT:
Code:
```x = (pi + 3) / 2

2x = pi + 3

2x(pi - 3) = (pi + 3)(pi - 3)

2pix - 6x = pi^2 - 9

9 - 6x = pi^2 - 2pix

9 - 6x + x^2 = pi^2 - 2pix + x^2

|(x - 3)|^2 = |(pi - x)|^2   <----from here down is changed

x - 3 = pi - x

2x=pi+3

pi = 2x-3
```
??
I don't know, I'm just guessing now.
This HAS to be right, I see nothing else that could be wrong.

Good thing there's no edit counter until someone posts after you, all my posts would say "edited 100 times", seriously. Almost every single post I make I go back and edit it multiple times.

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12 Aug 2009, 20:53
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
The other one:
The problem lies within x(x-1) = (x+1)(x-1), When you take that out it divides by zero..so you need to evaluate the expressions.

x^2-x=x^2-1 Simplify it -x=-1....x=1...1=1

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12 Aug 2009, 23:30
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
windwakr wrote:
The other one:
The problem lies within x(x-1) = (x+1)(x-1), When you take that out it divides by zero..
Yes, I thought that the division by zero had already been pointed out, but there is another piece to solve before the question as posed is resolved.
12 Aug 2009, 23:34
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
Ok, please start giving hints, I have nothing. It has been quite a long time since anyone other than me and sleepsleep have even attempted to answer these, we need some help.

I've completely mangled this one from it original formula, so its useless:
Code:
```                        1 = 1

x = 1

x = x

x-1 = x-1

x(x-1) = x(x-1)

x(x-1) = x²-x

x(x-1) = x²-1

x(x-1) = (x+1)(x-1)

x²-x   = x²-1    <--from here on is all changed

x²-x   = x²-1         ;0/-2=0
-----    -----
(-x-1)   (-x-1)

-x = -1

x = 1

1 = 1
```

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12 Aug 2009, 23:56
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
So, can we have a hint on the problems now, please?

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15 Aug 2009, 03:00
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
15 Aug 2009, 06:16
windwakr

Joined: 30 Jun 2004
Posts: 827
Location: Michigan, USA
windwakr
revolution wrote:
(3-x)^2 == (x-3)^2

Try it with any number you choose for x.

Ya, but when you take the square root of it like you do they are different:
Code:
``` (3 - x)^2 = (pi - x)^2

3 - x = pi - x

which reduces to:

3=pi
```

IS NOT EQUAL to:
Code:
``` (x - 3)^2 = (pi - x)^2

x - 3= pi - x

which reduces to:

2x=pi+3
```

I don't understand what you want by "make it correct.". Do you want it to still be pi=3 in the end?:
Code:
``` x = (pi + 3) / 2

2x = pi + 3

2x(pi - 3) = (pi + 3)(pi - 3)

2pix - 6x = pi^2 - 9

9 - 6x = pi^2 - 2pix

9 - 6x + x^2 = pi^2 - 2pix + x^2

|(3 - x)|^2 = |(pi - x)|^2

|3 - x| = |pi - x|

|pi| = |3|
```

Or what?

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15 Aug 2009, 14:32
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
windwakr wrote:
I don't understand what you want by "make it correct."
There is a way to express governing conditions in equations within the mathematical framework. My high school teacher would be very upset if the students did not properly express all the necessary information to ensure "mistakes" like this could not occur.
15 Aug 2009, 20:58
neville

Joined: 13 Jul 2008
Posts: 507
Location: New Zealand
neville
revolution wrote:
windwakr wrote:
I don't understand what you want by "make it correct."
There is a way to express governing conditions in equations within the mathematical framework. My high school teacher would be very upset if the students did not properly express all the necessary information to ensure "mistakes" like this could not occur.

The required "correction" is on the 2nd-to-last line:
Code:
```either - (3 - x) = pi - x
or    3 - x  = - (pi - x)    ```

This is just another case of deviously creating squares in the equation and then choosing to take only the positive square roots (a very old trick!)

In this case the only valid consistent solution is when square roots of the opposite sign are taken on each side of the equation (the 2 options given above, both of which confirm the original equation 2x = pi + 3.

When square roots of the same sign are taken (both +ve or both -ve) the result is pi=3 which is inconsistent with the original equation and therefore an invalid solution.

The other "mistake" in this example of Mickey Mouse Maths is to assume that the symbol "pi" in the equation has anything to do with the well-known mathematical constant (as strongly implied by you revolution!) In fact "pi" could be replaced with any other symbol and the algebra has nothing whatever to do with the definition of pi. In fact the algebra also proves absolutely nothing

I'll take my star now thanks revolution, although I think windwakr should receive at least a half-star so if you won't give him a full star too then I'll give him half of mine

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18 Aug 2009, 07:24
revolution
When all else fails, read the source

Joined: 24 Aug 2004
Posts: 17284
revolution
I would have also accepted
Code:
`±(3 - x) = (pi - x)    `
Or any other ASCII variants like
Code:
`+-(3 - x) = -+(pi - x)    `
Well done. I knew you guys would get it eventually. It was dangling on the edge for some time that it just had to come soon.

So anyhow, without further ado, I present the prize, with all the prestige that it deserves (i.e. none), to neville.

*

Congrats, and enjoy your fame while it lasts.

Now, let's see, there are still two open questions, who is willing to give them a try?

That's weird, the minus-plus character (∓, 0x2213) shows correctly in the edit box, but not in the main thread view in a code tag
18 Aug 2009, 15:52
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