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kohlrak
MHajduk wrote:
Merely a metaphor. 

24 Mar 2010, 11:10 

MHajduk
kohlrak wrote:


24 Mar 2010, 11:12 

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 ! 

24 Mar 2010, 11:19 

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.


24 Mar 2010, 11:45 

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 Code: a  c/2 = b  c/2 Code: a  c/2 = b  c/2 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 

24 Mar 2010, 11:54 

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. 

24 Mar 2010, 12:05 

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 squareroot notation without sign represents the positive square root, 

24 Mar 2010, 12:11 

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 

24 Mar 2010, 12:16 

revolution
I never showed any squareroot notation.
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. 

24 Mar 2010, 12:19 

revolution
MHajduk wrote: Your example should be solved this way: 

24 Mar 2010, 12:20 

MHajduk
revolution wrote: I never showed any squareroot notation. 

24 Mar 2010, 12:22 

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. 

24 Mar 2010, 12:25 

MHajduk
revolution wrote:
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):
Because relation of the implication is transitive we can write that
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. 

24 Mar 2010, 12:50 

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. 

24 Mar 2010, 12:56 

MHajduk
revolution wrote: Sure, no problem, sqrt(x^2) = x. But I never showed the sqrt(x^2) notation, that was added by you. sqrt = "square root" name of the function used in C 

24 Mar 2010, 13:02 

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.


24 Mar 2010, 13:07 

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. 

24 Mar 2010, 13:22 

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. 

24 Mar 2010, 13:27 

Borsuc
Why don't you just expand the parantheses to obtain a second degree polynomial?
_________________ Previously known as The_Grey_Beast 

24 Mar 2010, 13:58 

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