flat assembler
Message board for the users of flat assembler.

Index > Heap > Numerical oddities

Goto page Previous  1, 2, 3 ... 11, 12, 13 ... 24, 25, 26  Next
Author
Thread Post new topic Reply to topic
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
kohlrak wrote:
When all else fails, put in actual numbers... If the case is ever false, all it takes is one situation. Arguing with alphabet soup will get you nowhere.
Kohlrak, that's not "alphabet soup" that's algebra if you aren't awared yet.
Post 24 Mar 2010, 11:07
View user's profile Send private message Visit poster's website Reply with quote
kohlrak



Joined: 21 Jul 2006
Posts: 1421
Location: Uncle Sam's Pad
kohlrak
MHajduk wrote:
kohlrak wrote:
When all else fails, put in actual numbers... If the case is ever false, all it takes is one situation. Arguing with alphabet soup will get you nowhere.
Kohlrak, that's not "alphabet soup" that's algebra if you aren't awared yet.


Merely a metaphor. Razz
Post 24 Mar 2010, 11:10
View user's profile Send private message Visit poster's website AIM Address Yahoo Messenger MSN Messenger Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
kohlrak wrote:
MHajduk wrote:
kohlrak wrote:
When all else fails, put in actual numbers... If the case is ever false, all it takes is one situation. Arguing with alphabet soup will get you nowhere.
Kohlrak, that's not "alphabet soup" that's algebra if you aren't awared yet.


Merely a metaphor. Razz
Yeah, let it be. Wink
Post 24 Mar 2010, 11:12
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
So, revolution, are you still sure that my and yours answers are the same? Did you read this post http://board.flatassembler.net/topic.php?p=112372#112372?

You tried to "prove" that a = b for every a, b but we only could prove this way that

|a - b| = |b - a|

what is obvious for every a and b but from this you can't conclude that a = b !
Post 24 Mar 2010, 11:19
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
MHajduk: Nope. Hehe, it ain't that easy. You have changed the equation by adding the ||'s. It can be solved in a better way.
Post 24 Mar 2010, 11:45
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
I haven't "changed" the equation, you can't just "forget" about the second power here:
Code:
(a - c/2)^2 = (b - c/2)^2    
and write
Code:
a - c/2 = b - c/2    
you can write only something like this:
Code:
|a - c/2| = |b - c/2|    
OK, if you want it a bit simpler, you can write also this way:
Code:
             a + b = c 
    (a + b)(a - b) = c(a - b) 
         a^2 - b^2 = ca - cb 
          a^2 - ca = b^2 - cb 
a^2 - ca + (c^2)/4 = b^2 - cb + (c^2)/4 
       (a - c/2)^2 = (b - c/2)^2
         (a - b)^2 = (b - a)^2
         (a - b)^2 = (-(a - b))^ 2
         (a - b)^2 = (a - b)^2     (what, again, doesn't lead us to conclusion a = b !)
    


Last edited by MHajduk on 24 Mar 2010, 12:07; edited 1 time in total
Post 24 Mar 2010, 11:54
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
You can't simply say: "sqrt(x^2) = |x|"

How about this:

x=-3
x^2=9
sqrt(x^2)=|x| <---- your step is here.
x=3

It doesn't work like that. You follow a path that is not guaranteed.
Post 24 Mar 2010, 12:05
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Man, you don't know mathematics! See this article in Wikipedia (definition of absolute value) http://en.wikipedia.org/wiki/Absolute_value

Wikipedia wrote:
Since the square-root notation without sign represents the positive square root,
Image
which is sometimes even used as a definition of absolute value.
I only said that fault in your "reasoning" was when you "forget" about absolute value, hence any following conclusions were wrong.
Post 24 Mar 2010, 12:11
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
Your example should be solved this way:
Code:
x = -3
x^2 = 9
sqrt(x^2) = sqrt(9)
|x| = 3
hence
x = 3 or x = -3    
Post 24 Mar 2010, 12:16
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
I never showed any square-root notation. Wink

But you are right, I don't know much about mathematics, but I don't think you are safe to assume the absolute value in the equations I showed. And when you see the answer you will also see why your assumption about adding || is not best.
Post 24 Mar 2010, 12:19
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
MHajduk wrote:
Your example should be solved this way:
Code:
x = -3
x^2 = 9
sqrt(x^2) = sqrt(9)
|x| = 3
hence
x = 3 or x = -3    
So that shows that assuming that "sqrt(x^2) = |x|" is not correct. Right?
Post 24 Mar 2010, 12:20
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
revolution wrote:
I never showed any square-root notation. Wink

But you are right, I don't know much about mathematics, but I don't think you are safe to assume the absolute value in the equations I showed. And when you see the answer you will also see why your assumption about adding || is not best.
Again, in my first post I showed where we could correct your reasoning to obtain (obvious) result. This "trick" with "silent cancelling" of the second power is very old and well known but, of course, is not allowed in maths. Wink
Post 24 Mar 2010, 12:22
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
MHajduk wrote:
This "trick" with "silent cancelling" of the second power is very old and well known but, of course, is not allowed in maths. Wink
But it is allowed, and the answer, when you see it, will make it obvious. Try to think a little more about it. Your responses above are tantalising close to the solution. You just need to connect the dots.
Post 24 Mar 2010, 12:25
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
revolution wrote:
MHajduk wrote:
Your example should be solved this way:
Code:
x = -3
x^2 = 9
sqrt(x^2) = sqrt(9)
|x| = 3
hence
x = 3 or x = -3    
So that shows that assuming that "sqrt(x^2) = |x|" is not correct. Right?
No, sqrt(x^2) = |x| is true for all x.

Notice that if we transform equation we can widen the set of the solutions. We can't make transformations which may exclude some solutions of previous equations (as you tried to do with "cancelling" the second power).

Transformations presented above are correct because we widen the set of the possible solutions what is not wrong.

Let's denote above mentioned reasoning as a sequence of the implications (which are true):

  1. x = - 3 => x^2 = 9

    true, but we widen the set of the possible solutions from {-3} to {-3, 3}, this step is allowed (but not in the opposite direction)

  2. x^2 = 9 => sqrt(x^2) = sqrt(9)

    true, the set of the possible solutions {-3, 3}

  3. sqrt(x^2) = sqrt(9) => |x| = 3

    true, from the definition of the absolute value, set of possible solutions the same {-3, 3}

  4. |x| = 3 => [(x = - 3) or (x = 3)]

    true, again from the definition of the absolute value, set of possible solutions {-3, 3}


Because relation of the implication is transitive we can write that

  • x = -3 => [(x = -3) or (x = 3)]

    true, set of possible solutions {-3, 3}


We have an implication of the form

p => [p or q]

which is always true due to the properties of the implication.

Notice that (x = -3) or (x = 3) is equivalent to the fact that x belongs to the set {-3, 3}.

Well, what we obtain finally? We concluded that if x = -3 then x belongs to the set {-3, 3} what is absolutely true. Smile
Post 24 Mar 2010, 12:50
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
Sure, no problem, sqrt(x^2) = |x|. But I never showed the sqrt(x^2) notation, that was added by you.

There is another answer, a much better answer IMO.
Post 24 Mar 2010, 12:56
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
revolution wrote:
Sure, no problem, sqrt(x^2) = |x|. But I never showed the sqrt(x^2) notation, that was added by you.

There is another answer, a much better answer IMO.
I used notation of the square root possible to denote only with the Latin letters.

sqrt = "square root" name of the function used in C Wink
Post 24 Mar 2010, 13:02
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
I am not disputing that sqrt = square root. The issue is that you gave the notation, not me. I fully accept that sqrt meant square root. The Wikipedia page you showed supports your notation as given, but it was added by you so your result is true based upon something you added. In a more formal logic setting this is known as a "strawman argument". Hehe, but it sounds harsh to accuse you of that. So instead we can say based upon your new equation, your result it correct. But it is also not really relevant, because it changed what I wrote, rather than fixing what I wrote. I need something that fixes the equations I wrote.
Post 24 Mar 2010, 13:07
View user's profile Send private message Visit poster's website Reply with quote
MHajduk



Joined: 30 Mar 2006
Posts: 6034
Location: Poland
MHajduk
revolution wrote:
I am not disputing that sqrt = square root. The issue is that you gave the notation, not me. I fully accept that sqrt meant square root. The Wikipedia page you showed supports your notation as given, but it was added by you so your result is true based upon something you added. In a more formal logic setting this is known as a "strawman argument". Hehe, but it sounds harsh to accuse you of that. So instead we can say based upon your new equation, your result it correct. But it is also not really relevant, because it changed what I wrote, rather than fixing what I wrote. I need something that fixes the equations I wrote.
I think you've just changed rules of the game. Wink I was sure that we have to find where reasoning is wrong, not "fix" it. I've showed you how to correct your (improper) reasoning in two ways. Wink
Post 24 Mar 2010, 13:22
View user's profile Send private message Visit poster's website Reply with quote
revolution
When all else fails, read the source


Joined: 24 Aug 2004
Posts: 17279
Location: In your JS exploiting you and your system
revolution
I always just look for fixes, see the previous stuff. Admittedly I never explicitly stated it this time, so perhaps you have a rightful claim of a solution. So in that case I shall allow your result as a possible solution.

So now I formally correct my omission: Find a solution that fixes what I wrote. Where "fixes" means without introducing any new operations. MHajduk has already some good stuff that strongly hints at what the answer is.
Post 24 Mar 2010, 13:27
View user's profile Send private message Visit poster's website Reply with quote
Borsuc



Joined: 29 Dec 2005
Posts: 2466
Location: Bucharest, Romania
Borsuc
Why don't you just expand the parantheses to obtain a second degree polynomial? Confused

_________________
Previously known as The_Grey_Beast
Post 24 Mar 2010, 13:58
View user's profile Send private message Reply with quote
Display posts from previous:
Post new topic Reply to topic

Jump to:  
Goto page Previous  1, 2, 3 ... 11, 12, 13 ... 24, 25, 26  Next

< Last Thread | Next Thread >
Forum Rules:
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You can attach files in this forum
You can download files in this forum


Copyright © 1999-2020, Tomasz Grysztar. Also on YouTube, Twitter.

Website powered by rwasa.