True or False?
If x , y are real numbers such that x 2 = y 2 , then
x = y .
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If x 2 = a ⇒ x could be positive or negative.
If y 2 = a ( x 2 = y 2 ) ⇒ y could be positive or negative.
Therefore, it is true and false.
x 2 = y 2 means x 2 - y 2 = 0 so either x-y=0 (x=y) or x+y=0 (x=-y) so x and y are not always equal
the square root of x and y could be either positive or negative...
for example: 4=4
both x^2 and y^2 is 4
4 could have a square root of +2 and -2
if that's the case, +2 could be x and -2 could be y, which in that case, x=y won't be true
square root results in both positive and negative signs......... so not true always ....
in general , if x^2=y^2 . then x\quad =\quad \pm \sqrt { { y }^{ 2 } } therefore ,, x\quad =\quad \pm y
Sometimes true, sometimes false. √x²=√y², x=y, or, x=(-y), or, (-x)=y.
Counter example.
x = 2 and y = -2
So why not "Always false"?
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In case of x = 2 and y = 2, its true!!
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Right, so that needs to be stated in your solution. Almost all of us are not mind readers, and can only deduce from what you have written.
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@Calvin Lin – Sorry. I would take care of it.
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@Dev Sharma – I think the answer should be always false. Why? Well the question asks: True or false? and then makes a statement - an incorrect statement. It is indeed false that from x^2 = y^2, one cannot deduce x=y. It is possible with some other inputs, but using solely x^2 = y^2, that is impossible to deduce. So I would say that such a deduction is always logically incorrect.
The point is that I understood the question to be: "Is the following deduction always true, always false, or sometimes true and sometimes false?" and such a deduction is always false. Right?
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@Alija Bevrnja – @Calvin Lin , this reasoning is correct. Please check
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@Agnishom Chattopadhyay – The difference is the statement at the start, namely "If x and y are real numbers", which means that they are already fixed values. IE "If 2 and 2 are real numbers such that 2 2 = 2 2 , then 2 = 2 " is a true statement.
If the statement was x 2 = y 2 ⇒ x = y , where these are variables, then this is always false. In terms of variables, we can only say that x = ± y .
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@Calvin Lin – Okay, but how would an implication be stated using words. As far as I know, wording IF...THEN has the meaning of implication.
I do not see why the sentence "If x and y are real numbers..." has to imply fixed values. One has to say that they are real numbers, so that we know the domain of the problem...like writing x , y ∈ R .
What I am saying is that the full problem can be restated as: x , y ∈ R , x 2 = y 2 ⇒ x = y If not, then I ask the question, how to rephrase my mathematical statement (which can either be true or false - and is false in this case) into words.
I agree. Actually, a statement stated in general is False even if there is just one counterexample. I trust that logic here should be bivalent because the notion of 'not always' is ambiguous.
How about I say that x = 4x is sometimes true, because there is one solution?
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Consider 3 case:
(i) if x = y = 0 , we can have x 2 = y 2 = 0 . It means statement x 2 = y 2 ⟹ x = y always true.
(ii) if x < 0 , y > 0 ( or x > 0 , y < 0 ), we can have x 2 > 0 , y 2 > 0 . It means statement x 2 = y 2 ⟹ x = y always false, because the signs x , y are opposite.
(iii) if x < 0 , y < 0 ( or x > 0 , y > 0 ), we can have x 2 > 0 , y 2 > 0 . It means statement x 2 = y 2 ⟹ x = y always true, because the sign x , y are same.
From (i),(ii) and (iii) s o m e t i m e s t r u e , s o m e t i m e s f a l s e