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asd
#135 to #75

anon
Reply 0
(07/13/2013) [] It depends on whether one considers the principal square root function or the multivalued relation. Choosing between R and C is irrelevant (for what anyway? domain? codomain?). You can map from reals to reals with a real image and still have both the positives and negatives. It just won't be a function. R vs C is irrelevant.
#60 to #7

shredheadxd
Reply +1
(07/12/2013) [] If you start with 4, then take the square root, then you are right. The answers are 2, and 2.
But if you start with the square root of 4, then you are wrong. The only answer is 2.
Sincerely,
A 4th year math major, not that anybody cares.
But if you start with the square root of 4, then you are wrong. The only answer is 2.
Sincerely,
A 4th year math major, not that anybody cares.
#137 to #60

anon
Reply 0
(07/13/2013) [] Not really. In this case, it's not really clear whether or not we're talking about the just the principal value or not. Unlike written out mathematical notation, words can be ambiguous on this matter. So we clarify,
You start with 4, take the (principal) square root, and you will only have 2. Likewise, if you start with the (principal) square root of 4, then you will still only have 2.
You start with 4, take the (principal) square root, and you will only have 2. Likewise, if you start with the (principal) square root of 4, then you will still only have 2.
#143 to #137

shredheadxd
Reply +1
(07/13/2013) [] On the contrary,
When you do the action of square rooting, you must consider the positive and negative cases. Since you do not know whether the square root was principal or not, you must consider both cases. The positive case is the one shown in the content  the principal square root. The negative case will give you 2.
If you start with simply "square root of 4", you have skipped the action of square rooting, and jump right to evaluating the positive case (the principal square root).
It is quite extremely crystal clear.
When you do the action of square rooting, you must consider the positive and negative cases. Since you do not know whether the square root was principal or not, you must consider both cases. The positive case is the one shown in the content  the principal square root. The negative case will give you 2.
If you start with simply "square root of 4", you have skipped the action of square rooting, and jump right to evaluating the positive case (the principal square root).
It is quite extremely crystal clear.
#204 to #202

shredheadxd
Reply 0
(07/14/2013) [] Lemme break it down.
±√4 = +√4 AND √4 (it's "AND" or "OR" depending on the situation)
+√4 = √4 = 2
√4 = 2
So if you just look at the above statements, you can see that √4 = 2. Not 2. Math don't lie!
±√4 = +√4 AND √4 (it's "AND" or "OR" depending on the situation)
+√4 = √4 = 2
√4 = 2
So if you just look at the above statements, you can see that √4 = 2. Not 2. Math don't lie!
#206 to #205

shredheadxd
Reply 0
(07/15/2013) [] that's correct! I wasn't sure what you were talking about, but the math wasn't ambiguous so I just reiterated it.
#54 to #7

mrloverlover
Reply +15
(07/12/2013) [] everyone is replying to this with math equations etc.
and i'm just sat here thinking what the **** is wrong with that dog
and i'm just sat here thinking what the **** is wrong with that dog
#33 to #7

bronybox
Reply +4
(07/12/2013) [] Get ready to learn ************(s).
As displayed by the graph of a square root function (shown on the left), there are no negative y values, and thus f(x) can never be negative [ sqrt(2) can never equal  2 ]. This is because the square root operation IS A function, and thus their can only be 1 y value for every x value (but not necessarily the other way around).
You are stating one of the most common misconceptions people have once they are taught to solve quadratic equations, observe.
1. x^2 = 4
2. x = sqrt (4)
3. x = ± 2
You make the assumption that the square root of 4 is equal to +/ 2, however, by actual mathematical definition, line 2 holds a mistake. This is because x DOES NOT equal sqrt(4). The correct solution is:
1. x^2 = 4
2. x = ± sqrt(4)
3. x = ± 2
Just as you cannot take the square root of a negative number, the square root of a positive number can never be negative. The radical symbol by definition is only the principal square root, i.e., always positive.
As displayed by the graph of a square root function (shown on the left), there are no negative y values, and thus f(x) can never be negative [ sqrt(2) can never equal  2 ]. This is because the square root operation IS A function, and thus their can only be 1 y value for every x value (but not necessarily the other way around).
You are stating one of the most common misconceptions people have once they are taught to solve quadratic equations, observe.
1. x^2 = 4
2. x = sqrt (4)
3. x = ± 2
You make the assumption that the square root of 4 is equal to +/ 2, however, by actual mathematical definition, line 2 holds a mistake. This is because x DOES NOT equal sqrt(4). The correct solution is:
1. x^2 = 4
2. x = ± sqrt(4)
3. x = ± 2
Just as you cannot take the square root of a negative number, the square root of a positive number can never be negative. The radical symbol by definition is only the principal square root, i.e., always positive.
#195 to #169

kikisu
Reply 0
(07/13/2013) [] I am. If you take the root of 4, its both 2 and 2. But if you already have the square root of four, its just two. I do understand the definition of a square root, with is a number that when multiplied by itself gives you the original. 2 times 2 is 4.
#199 to #198

bronybox
Reply 0
(07/13/2013) [] But when you take the square root, you are taking the positive and negative root, and you SHOULD technically be using the notation "±sqrt (a)", otherwise what you are writing is simply incorrect by means of notation. You may be doing the right thing in your head, but you're not writing down what you should be.
#134 to #119

anon
Reply 0
(07/13/2013) [] The only bad part I see is "the square root of a positive number can never be negative". However, given the context one can easily tell that he's still referring to the principal square root.
Besides this, how is he incorrect? (unless this is what you meant)
Besides this, how is he incorrect? (unless this is what you meant)
#30 to #29

yunablade
Reply 2
(07/12/2013) [] Maybe it's a cultural thing, when I took math I was required to express the results to a square root always as +/ (unless imaginary number) and if the root was part of a bigger operation record both results (when the value is negative and when positive)
#36 to #24

bronybox
Reply 1
(07/12/2013) [] You are 100% correct, there are two solutions to x^2 = 4, ±2
HOWEVER.
You are wrong in the fact that √x² = √4
It should be ±√x² = ±√4
Because √x is a function (let's denote it f(x) = √x)
f(x) can never equal both 2 and 2 when x = 4 by the definition of a function.
HOWEVER.
You are wrong in the fact that √x² = √4
It should be ±√x² = ±√4
Because √x is a function (let's denote it f(x) = √x)
f(x) can never equal both 2 and 2 when x = 4 by the definition of a function.