The square root yields only one value ?

In summary, the conversation discussed the idea of square roots yielding only positive values and the proof of this concept. It was mentioned that the question "What is the value of x, if x^2=4?" has two solutions, +2 and -2, whereas asking for the square root of 4 only has one solution, 2. The conversation also touched on the definition of complex numbers and the importance of distinguishing between i and -i. Ultimately, the conclusion was that the proof given was incorrect and the concept of square roots yielding only positive values is not always true.
  • #1
sankalpmittal
785
15
The square root yields only one value !?

One guy advanced and asked me the question with an intention to judge my skill - " What is the value of x , if x2=4 " I , without any reluctance replied rather confidently - " Ha , its simple x will have 2 values : +2 and -2 "

He laughed hard and said " Don't you know , the value of x will be only positive ie +2 . Square root never yield negative values ie 41/2 = 2 and not -2 . "His monotonous and disdained voice made me feel uneasy . I replied " How can you prove it ? "He then gave me the following theorems :

1. To prove roots yield only positive values :1=1x1
1=-1x-1
So ;
1x1=-1x-1
On square root both sides :
(1x1)1/2 =(-1x-1)1/2
1=i2
1=-1 ?

or

http://www.artofproblemsolving.com/Wiki/index.php/Square_root [Broken]

2. Because of the discontinuous nature of the square root function in the complex plane, the law √zw = √z√w is in general not true. (Equivalently, the problem occurs because of the freedom in the choice of branch. The chosen branch may or may not yield the equality; in fact, the choice of branch for the square root need not contain the value of √z√w at all, leading to the equality's failure. A similar problem appears with the complex logarithm and the relation log z + log w = log(zw).) Wrongly assuming this law underlies several faulty "proofs", for instance the following one showing that –1 = 1:

http://en.wikipedia.org/wiki/Square_root#Algebraic_formula

3.
√-1/√-2 does not equal to √(-1/-2) if nos < 1 .
Sorry I forgot what he gave for this .

_____________________________________________________________________

Can anyone draw conclusion for this conversation ? Do roots really yield only positive values ? Please anyone illustrate these theorems more comprehensively .
:(

Thanks :)

( I am 14 years , class or year 10 . )
 
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  • #2


Hmm... He said x2 = 4.

Essentially, it is y = x2 where y = 4. If you graph y = x2, there are two places where y=4, as you said - 2 and -2. So, technically, he is incorrect there for that particular question.

If, on the other hand, he said x = 41/2, then he may have a case there. Didn't read over his proof that much. If I were to make a comment on the proofs, I may possibly make a mistake. Sorry. :frown:
 
  • #3


"What is the value of x if [itex]x^2= 4[/itex]?" and "What is the square root of 4?" are two completely different questions. Believing that [itex]x^2= 4[/itex] has only x= 2 as solution is an example of "A little learning is a dangerous thing"! ("Drink deep, or taste not the Pierian spring". Alexander Pope)

By definition, for a any positive number, [itex]a^{1/2}= \sqrt{a}[/itex] is the positive number satisfying [itex]x^2= a[/itex] but that does not mean that the equation itself does not have two solutions! Indeed the reason we need the "[itex]\pm[/itex]" in writing the solution as [itex]x= \pm\sqrt{a}[/itex] is that [itex]\sqrt{a}[/itex] along gives only one of them.
 
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  • #4


x2=4

This yields 2 or -2.
However, if he were to ask 41/2, it would be two since it is defined as the absolute value.
 
  • #5


Others have already explained what is going on. But this caught my eye.

sankalpmittal said:
(1x1)1/2 =(-1x-1)1/2
1=i2
1=-1 ?

This is nonsense. i is defined as i2=-1. Do you really think some random nobody has managed disproved 100 years of complex analysis?
 
  • #6


Man, knew there was something wrong when I saw it, but I probably didn't want to comment on it.
 
  • #7


pwsnafu said:
Others have already explained what is going on. But this caught my eye.



This is nonsense. i is defined as i2=-1. Do you really think some random nobody has managed disproved 100 years of complex analysis?

i can't be defined that way, as that definition does not distinguish i from -1.
 
  • #8


SteveL27 said:
i can't be defined that way, as that definition does not distinguish i from -1.

You mean i from -i right, and not i from -1? Yes I know, but the full construction via polynomial rings would go right over OP's head. The defining property of complex numbers is still i2=-1, as opposed to other systems.
 
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  • #9


pwsnafu said:
You mean i from -i right, and not i from -1? Yes I know, but the full construction via polynomial rings would go right over OP's head. The defining property of complex numbers is still i2=-1, as opposed to other systems.

Yes -i, thx.
 
  • #10


hubewa said:
Hmm... He said x2 = 4.

Essentially, it is y = x2 where y = 4. If you graph y = x2, there are two places where y=4, as you said - 2 and -2. So, technically, he is incorrect there for that particular question.

If, on the other hand, he said x = 41/2, then he may have a case there. Didn't read over his proof that much. If I were to make a comment on the proofs, I may possibly make a mistake. Sorry. :frown:

HallsofIvy said:
"What is the value of x if [itex]x^2= 4[/itex]?" and "What is the square root of 4?" are two completely different questions. Believing that [itex]x^2= 4[/itex] has only x= 2 as solution is an example of "A little learning is a dangerous thing"! ("Drink deep, or taste not the Pierian spring". Alexander Pope)

By definition, for a any positive number, [itex]a^{1/2}= \sqrt{a}[/itex] is the positive number satisfying [itex]x^2= a[/itex] but that does not mean that the equation itself does not have two solutions! Indeed the reason we need the "[itex]\pm[/itex]" in writing the solution as [itex]x= \pm\sqrt{a}[/itex] is that [itex]\sqrt{a}[/itex] along gives only one of them.

pwsnafu said:
Others have already explained what is going on. But this caught my eye.



This is nonsense. i is defined as i2=-1. Do you really think some random nobody has managed disproved 100 years of complex analysis?

SteveL27 said:
i can't be defined that way, as that definition does not distinguish i from -1.

pwsnafu said:
You mean i from -i right, and not i from -1? Yes I know, but the full construction via polynomial rings would go right over OP's head. The defining property of complex numbers is still i2=-1, as opposed to other systems.

SteveL27 said:
Yes -i, thx.

BloodyFrozen said:
x2=4

This yields 2 or -2.
However, if he were to ask 41/2, it would be two since it is defined as the absolute value.

hubewa said:
Man, knew there was something wrong when I saw it, but I probably didn't want to comment on it.

Hmm , can anyone illustrate the theorems with more clarity . I searched on google and found that the "guy" was correct !
:confused:
 
  • #11


sankalpmittal said:
Hmm , can anyone illustrate the theorems with more clarity . I searched on google and found that the "guy" was correct !
:confused:

Sure. This will be long and heavy. Read it in bite sized chunks.:smile:

Firstly, what I write in this post is about the real numbers only. It does not generalize to other number systems. Double emphasis is required. Your friend mucked this up.

Okay, in algebra we classify number systems by their properties and give names to them. One of the most important is what we call a http://en.wikipedia.org/wiki/Field_%28mathematics%29" [Broken]. It is a set with addition, subtraction, multiplication and division-by-nonzero. Examples include the rational numbers (fractions), real numbers and complex numbers, and others.

For this discussion we need the concept of an http://en.wikipedia.org/wiki/Ordered_field" [Broken]. This means that we have the concept of "less than" and further if x >y, then for any number a, we have x+a > y+a. This is an important property of real numbers.

Now, consider the following problem:
Let a be a non-negative real number. For what x, do we have x2=a?
Clearly, this is the problem your friend gave you.
Answer: If a=0, then there only exists one solution, namely x equal to zero. If a is not zero, then there exist two solutions.

Why two? Well, suppose that we call x1 to be one solution. Then let x2 = -x1. We see that

x22 = (-x1)2 = x12
= a,

and we have a second solution. This is the "two solutions" answer you gave to your friend. You are correct in this.

Now, remember, the reals are an ordered field. Thus one of the solutions is greater than the other. It may be x1, or x2. We don't know. But we do know that one is greater than zero, and the other is less than zero. Why do we know this? Because we chose x2 = -x1.

This means we can distinguish the two. We now define the notation [itex]\sqrt{a}[/itex] to be the positive solution. So if I ask you "What is the square root of a?", the correct solution is [itex]\sqrt{a}[/itex]. This is not the same question as the previous one. I cannot stress that enough. In the latter, we are specifically asking for the positive solution, so there is only one solution. In the former we are asked to solve a quadratic equation, so there are two solutions.

Diversion: Before we continue, sankalpmittal, how much do you know about complex numbers?
 
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  • #12


I was just rereading the OP.

sankalpmittal said:
One guy advanced and asked me the question with an intention to judge my skill - " What is the value of x , if x2=4 " I , without any reluctance replied rather confidently - " Ha , its simple x will have 2 values : +2 and -2 "

He laughed hard and said " Don't you know , the value of x will be only positive ie +2 . Square root never yield negative values ie 41/2 = 2 and not -2 . "

Wrong rebuttal! Correct rebuttal is "Prove that -2 does not square to 4." Unfortunately, you have fallen to a http://en.wikipedia.org/wiki/Strawman" [Broken]. The original problem was about quadratic equations not the square root symbol nor the square root function (which is what the art of problem solving link is on about). He has deflected your answer by bringing something unrelated to the problem.

2. Because of the discontinuous nature of the square root function in the complex plane, the law √zw = √z√w is in general not true. (Equivalently, the problem occurs because of the freedom in the choice of branch. The chosen branch may or may not yield the equality; in fact, the choice of branch for the square root need not contain the value of √z√w at all, leading to the equality's failure. A similar problem appears with the complex logarithm and the relation log z + log w = log(zw).) Wrongly assuming this law underlies several faulty "proofs", for instance the following one showing that –1 = 1:

http://en.wikipedia.org/wiki/Square_root#Algebraic_formula

This proves you are right and your friend is wrong! The formula you cite is
[itex]\sqrt{x+iy} = \sqrt{(x+r)/2}\pm\sqrt{(x-r)/2}[/itex]
This means that the left hand side has two solutions, and so is a direct counterexample to your friend's assertion that the square root yields only one solution. In context, he means that the square root of a non-negative real number yields only one solution, but considering he was the one who brought complex numbers into the discussion as proof, it's his fault.


( I am 14 years , class or year 10 . )

Bah! I should have read that first! I thought this was a sig, and I usually don't read sigs.
So I can assume you know next to nothing about complex numbers, right?
 
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  • #13


HallsofIvy said:
"What is the value of x if [itex]x^2= 4[/itex]?" and "What is the square root of 4?" are two completely different questions.
Those are the same questions. The different question would be "What is the value of x if [itex]x = \sqrt{4}[/itex]" or "What is the principle square root of 4?".

"What is the square root of 4?"
- The square-root of 4 is the number that was squared to produce 4. It could be 2 or it could be -2. So the only conclusion I can make is, "the square-root of 4 is 2 or -2."
 
  • #14


HallsofIvy said:
"What is the value of x if [itex]x^2= 4[/itex]?" and "What is the square root of 4?" are two completely different questions. .

GenePeer said:
Those are the same questions. The different question would be "What is the value of x if [itex]x = \sqrt{4}[/itex]" or "What is the principle square root of 4?".

"What is the square root of 4?"
- The square-root of 4 is the number that was squared to produce 4. It could be 2 or it could be -2. So the only conclusion I can make is, "the square-root of 4 is 2 or -2."
No, in the real number system, which is what is being used here, [itex]\sqrt{x}[/itex], square root, is a function and has only value for each x. [itex]2= \sqrt{4}[/itex], not just the "principle root".

Would you agree that the solution to the equation [itex]x^2= a[/itex] is [itex]x= \pm\sqrt{a}[/itex]?

If [itex]\sqrt{a}[/itex] is both positive and negative, why do we need that "[itex]\pm[/itex]"?
 
  • #15


I don't think we can have a square-root function because by definition square-roots have double-values. So [itex]\sqrt{x}[/itex] refers to the principle square-root which can indeed be a function.
 
  • #16


pwsnafu said:
Sure. This will be long and heavy. Read it in bite sized chunks.:smile:

Firstly, what I write in this post is about the real numbers only. It does not generalize to other number systems. Double emphasis is required. Your friend mucked this up.

Okay, in algebra we classify number systems by their properties and give names to them. One of the most important is what we call a http://en.wikipedia.org/wiki/Field_%28mathematics%29" [Broken]. It is a set with addition, subtraction, multiplication and division-by-nonzero. Examples include the rational numbers (fractions), real numbers and complex numbers, and others.

For this discussion we need the concept of an http://en.wikipedia.org/wiki/Ordered_field" [Broken]. This means that we have the concept of "less than" and further if x >y, then for any number a, we have x+a > y+a. This is an important property of real numbers.

Now, consider the following problem:

Clearly, this is the problem your friend gave you.
Answer: If a=0, then there only exists one solution, namely x equal to zero. If a is not zero, then there exist two solutions.

Why two? Well, suppose that we call x1 to be one solution. Then let x2 = -x1. We see that

x22 = (-x1)2 = x12
= a,

and we have a second solution. This is the "two solutions" answer you gave to your friend. You are correct in this.

Now, remember, the reals are an ordered field. Thus one of the solutions is greater than the other. It may be x1, or x2. We don't know. But we do know that one is greater than zero, and the other is less than zero. Why do we know this? Because we chose x2 = -x1.

This means we can distinguish the two. We now define the notation [itex]\sqrt{a}[/itex] to be the positive solution. So if I ask you "What is the square root of a?", the correct solution is [itex]\sqrt{a}[/itex]. This is not the same question as the previous one. I cannot stress that enough. In the latter, we are specifically asking for the positive solution, so there is only one solution. In the former we are asked to solve a quadratic equation, so there are two solutions.

Diversion: Before we continue, sankalpmittal, how much do you know about complex numbers?


Hmmm , I think that the "guy's" proofs have been mixed up . In the first theorem he illustrates a step that (-1x-1)1/2 (In LHS) = -11/2x-11/2 or i2 . But generally we consider that a1/2xb1/2 is not equal to (axb)1/2 , assuming that a and b both are less than 0 .

pwsnafu said:
I was just rereading the OP.



Wrong rebuttal! Correct rebuttal is "Prove that -2 does not square to 4." Unfortunately, you have fallen to a http://en.wikipedia.org/wiki/Strawman" [Broken]. The original problem was about quadratic equations not the square root symbol nor the square root function (which is what the art of problem solving link is on about). He has deflected your answer by bringing something unrelated to the problem.



This proves you are right and your friend is wrong! The formula you cite is
[itex]\sqrt{x+iy} = \sqrt{(x+r)/2}\pm\sqrt{(x-r)/2}[/itex]
This means that the left hand side has two solutions, and so is a direct counterexample to your friend's assertion that the square root yields only one solution. In context, he means that the square root of a non-negative real number yields only one solution, but considering he was the one who brought complex numbers into the discussion as proof, it's his fault.




Bah! I should have read that first! I thought this was a sig, and I usually don't read sigs.
So I can assume you know next to nothing about complex numbers, right?


I am in class 10th but still I know little about complex numbers . Complex numbers are the universal set of all existing numbers . They are divided into 2 : Real numbers and imaginary or unreal numbers .

Imaginary numbers are numbers in which the discriminant ie b2-4ac<0.
In real numbers b2-4ac >0 or b2-4ac = 0 .
:)

I don't know why was that guy messing up real nos question with complex group ?!
Certain sites it is clearly illustrated that in quadratic equations root can have two values but if we say square root of a number say 4 , or only roots then it will yield only 2 as solution not -2 .

:confused:
 
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  • #17


sankalpmittal said:
Hmmm , I think that the "guy's" proofs have been mixed up . In the first theorem he illustrates a step that (-1x-1)1/2 (In LHS) = -11/2x-11/2 or i2 . But generally we consider that a1/2xb1/2 is not equal to (axb)1/2 , assuming that a and b both are less than 0 .




I am in class 10th but still I know little about complex numbers . Complex numbers are the universal set of all existing numbers .
The complex numbers is the largest set of numbers commonly used but it is possible (and has been done) to define extensions to the complex numbers, such as the "quaternions".

They are divided into 2 : Real numbers and imaginary or unreal numbers .
No. there are complex numbers that are combinations of real and imaginary numbers.

Imaginary numbers are numbers in which the discriminant ie b2-4ac<0.
In real numbers b2-4ac >0 or b2-4ac = 0 .
That makes no sense because you haven't said what a, b, and c are! I assume you are talking about solutions to the quadratic equation [itex]ax^2+ bx+ c= 0[/itex]. That is, of course, only one way out of many of working with complex numbers.

:)

I don't know why was that guy messing up real nos question with complex group ?!
Certain sites it is clearly illustrated that in quadratic equations root can have two values but if we say square root of a number say 4 , or only roots then it will yield only 2 as solution not -2 .

:confused:
 
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  • #18


Well, to be honest, that's what we have been saying all along.

There are fundamental differences between these two functions

y = x2

and

y1/2 = x.

and what you have stated in your last post about this is correct.

PS To other ppl who have been using LaTeX, how do you use it on this forum?
 
  • #19


hubewa said:
Well, to be honest, that's what we have been saying all along.

There are fundamental differences between these two functions

y = x2

and

y1/2 = x.

and what you have stated in your last post about this is correct.
Who's post are you referring to?

PS To other ppl who have been using LaTeX, how do you use it on this forum?
use [ tex ] and [ \tex ] (without the spaces) or [ itex ] [ \itex ] for "in line" formulas (again without the spaces.

There is, currently, a mildly annoying problem with the LaTeX- you may have to click on the "refresh" button on your browser for it to work.
 
  • #20


Who's post are you referring to?

I think we posted at the same time. Maybe I should use quotes more.

I was referring to sank's latest post.

Anyway, thanks for that info about LaTeX.
 
  • #21


You cannot find the square root of a negative number without invoking i. You can have a negative number as a possible root of a positive number.
 
  • #22


Jack23454 said:
You cannot find the square root of a negative number without invoking i.

I hear the physics majors do just fine without i. They call it j.:smile:
 
  • #23


SteveL27 said:
I hear the physics majors do just fine without i. They call it j.:smile:
Good Point! At Haifa we used i, but it was a long time ago.
 
  • #24


Actually, that's more an engineering notation than phyiscs.
 
  • #25


HallsofIvy said:
Actually, that's more an engineering notation than phyiscs.

My most humble apologies to the physicists!
 

1. What is the definition of a square root?

A square root is a number that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because 3 multiplied by itself equals 9.

2. How do you calculate the square root of a number?

To calculate the square root of a number, you can use a calculator or a mathematical formula. The most common way to calculate the square root is by using the square root symbol (√) or by raising the number to the power of 1/2.

3. Does every number have a square root?

Yes, every positive number has a square root. However, some numbers have irrational square roots, meaning they cannot be expressed as a simple fraction or decimal. For example, the square root of 2 is an irrational number.

4. Can a negative number have a square root?

No, a negative number cannot have a real square root. This is because when a negative number is squared, it becomes positive. However, imaginary numbers can have square roots, such as the square root of -1 being the imaginary number i.

5. Why does the square root only yield one value?

A square root only yields one value because it is the inverse operation of squaring a number. When we square a number, we are multiplying it by itself, so there is only one possible solution to finding the original number. Similarly, when we take the square root, we are finding the number that was squared, so there is only one possible solution.

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