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We might start by taking simple and obvious identity:
1 = 1
this implies 13 - 12 = 13 - 12
or 9 - 12 + 4 = 4 - 12 + 9
or 3^2 - 2*3*2 + 2^2 = 2^2 - 2*2*3 + 3^2
or (3-2)^2 = (2-3)^2
taking the square roots on the both sides, we have
3-2 = 2-3
Rearranging the terms we have
6 = 4
or 3 = 2
deducting 1 from each sides we have
2 = 1
Now multiplying both the sides by 4 we have
8 = 4 = 2 + 2
Q.E.D
> or (3-2)^2 = (2-3)^2
> taking the square roots on the both sides, we have
> 3-2 = 2-3
This step is wrong. Any positive number has two square roots: one positive and one negative.
When we refer to "the" square root, we refer to the positive root only (by convention).
you are contradicting your own two statements. Once you have said that "any positive number has two square roots", how come you make your 'own convention' of taking only the positive one?
Example: if x^2 = 4 then
x = 2 and x = -2 both are valid solutions. If you feel the statement "taking the square roots on bothe the sides", too ambigious, substitute it by the statement "removing the radicals on both the sides". That will remove any ambiguity in the statement.
This step is mathematically incorrect.Quote:
Originally Posted by mahadev2
Ifthen
is correct.
(x^2)=(-x)^2 Your error has been to say that it follows that x=-x In this case the only solution would be x.
You had (3-2)^2=(2-3)^2 That is (1)^2=(-1)^2 which is true but it doesn't follow that 1=-1
How do you intend to remove the radicals on both sides?
In my last reply it should have said at end of 1st line 'only solution would be x=0'
It doesn't adds anything new to your 'argument' which is clearly a tautology simply because you are starting at the end result itself (x = - x) which is not the case with my derivation which starts from a very simple and obvious stage. Because of the limitation of the tool that i'm using, i had to use that phrase 'radical sign'. Otherwise it is very obvious what my intention is.
That's why I like mathematics. Unlike in some other areas of human thought, mathematical reasoning can be checked by machines, so that (at least in simple cases) there is no doubt if a certain proof is correct. Leibniz hoped to achieve the same level of clarity in other areas:
@OP: As was said above, the statementQuote:
Originally Posted by Gottfried Leibniz
For all real numbers x and y, if x2 = y2, then x = y
which you use after the words "taking the square roots on the both sides," is false. If you claim otherwise, please exhibit a derivation of this fact, e.g., from the axioms of real numbers and preferably using mostly symbols, not words.
as to 1st statement - your values are +ve nos! Cheerios is not appling his own convention.
Secondly - replace (3-2)^2 and (2-3)^2 with 1*1 and -1*-1 - this is still valid.
now you can simplfy further by dividing by -1 or 1 either way you will get -1=-1 . this avoids the whole square root issue if you dont understand it and also invalidates the proof.
> as to 1st statement - your values are +ve nos! Cheerios is not appling his own convention.
I don't know what you are referencing. If I ever refer to the square root (singular) I will be talking about the positive root...though I would rather use more precise language.
> Once you have said that "any positive number has two square roots", [from mahadev2]
O agree with your analysis and approach.... just trying to back you up
Then neither of you compehends the difference between "two square roots" and "the square root". Those are two completely different statements.
so enlighten us afterall this is a help forum
I will repeat what I posted in #4.Quote:
Originally Posted by Mytutoing
The square root ofis
Thus the square root of, That was the mistake in the beginning.