The Square Root of Numbers

In summary: The first root is what you get when you solve for x, and the second root is what you get when you solve for -x. So when you have an equation like x^2=-5, the two solutions are x=-5 and x=sqrt(-5).
  • #1
frozonecom
63
0

Homework Statement



What I am confused about is:
[itex] x - \sqrt{2}=0[/itex]
[itex]x=\sqrt{2}[/itex]
is it equal to
[itex]x=\pm\sqrt{2}[/itex]

?

Homework Equations


The Attempt at a Solution



The way I know it, those two are not equal. But my teacher insists on saying that whenever there is a square root, we should always recognize also the negative root.

It says based from this post : https://www.physicsforums.com/showpost.php?p=3618911&postcount=2 that it isn't.

Isn't it you only consider to get the positive and negative root if it's like this:
[itex]x^2=25[/itex]
Then that's when you say that [itex]x=\pm5[/itex]

Is there any law or postulate that clears this situation??
Please correct me if I am wrong and if my teacher is actually correct.

And if I am correct, how can I explain it to her so that she'll get it right away? I'm not that good at explaining anyway.
 
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  • #2
frozonecom said:

Homework Statement



What I am confused about is:
[itex] x - \sqrt{2}=0[/itex]
[itex]x=\sqrt{2}[/itex]
is it equal to
[itex]x=\pm\sqrt{2}[/itex]

?

Homework Equations





The Attempt at a Solution



The way I know it, those two are not equal. But my teacher insists on saying that whenever there is a square root, we should always recognize also the negative root.

It says based from this post : https://www.physicsforums.com/showpost.php?p=3618911&postcount=2 that it isn't.

Isn't it you only consider to get the positive and negative root if it's like this:
[itex]x^2=25[/itex]
Then that's when you say that [itex]x=\pm5[/itex]

Is there any law or postulate that clears this situation??
Please correct me if I am wrong and if my teacher is actually correct.

And if I am correct, how can I explain it to her so that she'll get it right away? I'm not that good at explaining anyway.


how is: [itex]x=\pm5[/itex]

different from: [itex]x=\pm \sqrt{25}[/itex]
?
 
  • #3
Zula110100100 said:
how is: [itex]x=\pm5[/itex]

different from: [itex]x=\pm \sqrt{25}[/itex]
?

No, what I meant in my post is when an equation is not quadratic, as :

[itex]x-\sqrt{25}=0[/itex]

is different from when the equation is quadratic:

[itex]x^2-25=0[/itex]

So, the solution for the quadratic equation is:
[itex]x=\pm5[/itex]

I'm asking if the solution for the first equation will be:
[itex]x=\pm\sqrt{25}[/itex]
or is it just the positive one??

If so, what is the explanation?
 
  • #4
I am perhaps not the best to answer this, but in my opinion(which doesn't count a lot in math) It could be either, because -5^2 = 25, so i we use -5, x+5 =0, x = -5, if we +5 we have x-5 = 0, x = 5

You could just write [itex] x = \sqrt{25} = \pm 5[/itex]
 
  • #5
Hi frozonecom! :smile:

Obviously
[itex]x=\sqrt{2}[/itex]
is not the same as
[itex]x=\pm\sqrt{2}[/itex]But if you solve
[itex]x^2=25[/itex]
the solution is
[itex]x=\pm\sqrt{25}[/itex]Note that the square root function of a positive real number is defined to be always positive.
 
  • #6
frozonecom said:
No, what I meant in my post is when an equation is not quadratic, as :

[itex]x-\sqrt{25}=0[/itex]

is different from when the equation is quadratic:

[itex]x^2-25=0[/itex]

So, the solution for the quadratic equation is:
[itex]x=\pm5[/itex]

I'm asking if the solution for the first equation will be:
[itex]x=\pm\sqrt{25}[/itex]
or is it just the positive one??

If so, what is the explanation?

Can it be the minus ? Think. What does the solution of an equation means ? If you plug the solution back to the equation, you should get something like 0=0, or other trivial equality.

So x-√25=0. How do you solve this equation ? What is the solution ?
 
  • #7
frozonecom said:
What I am confused about is:
[itex] x - \sqrt{2}=0[/itex]
[itex]x=\sqrt{2}[/itex]
is it equal to
[itex]x=\pm\sqrt{2}[/itex]
No.
If [itex] x - \sqrt{2}=0[/itex], then [itex] x = \sqrt{2}[/itex]. Period.
x is a positive number, approximately 1.414.
 
  • #8
frozonecom said:

Homework Statement



What I am confused about is:
[itex] x - \sqrt{2}=0[/itex]
[itex]x=\sqrt{2}[/itex]
is it equal to
[itex]x=\pm\sqrt{2}[/itex]

?

Homework Equations





The Attempt at a Solution



The way I know it, those two are not equal. But my teacher insists on saying that whenever there is a square root, we should always recognize also the negative root.

It says based from this post : https://www.physicsforums.com/showpost.php?p=3618911&postcount=2 that it isn't.

Isn't it you only consider to get the positive and negative root if it's like this:
[itex]x^2=25[/itex]
Then that's when you say that [itex]x=\pm5[/itex]

Is there any law or postulate that clears this situation??
Please correct me if I am wrong and if my teacher is actually correct.

And if I am correct, how can I explain it to her so that she'll get it right away? I'm not that good at explaining anyway.

You need to distinguish between "root of a number" and the "square-root *function*". Every piece of software I know of treats the square-root _function_ (sqrt or sqr or ...) as a single POSITIVE value (provided that you start with a positive value whose root you want). However, there are two *roots* to the equation x^2 = a (a > 0), namely: x = sqrt(a) and x = -sqrt(a). '

RGV
 
  • #9
frozonecom,

Only when you take the square root of something do you put ±. If there already is a square root, leave it as it is.
 
  • #10
Zula110100100 said:
I am perhaps not the best to answer this, but in my opinion(which doesn't count a lot in math) It could be either, because -5^2 = 25, so i we use -5, x+5 =0, x = -5, if we +5 we have x-5 = 0, x = 5
I'm no expert, but I do know that the bolded is not correct. -5^2 = -25. Do you see why?
 
  • #11
Zula110100100 said:
I am perhaps not the best to answer this, but in my opinion(which doesn't count a lot in math) It could be either, because -5^2 = 25, so i we use -5, x+5 =0, x = -5, if we +5 we have x-5 = 0, x = 5
eumyang said:
I'm no expert, but I do know that the bolded is not correct. -5^2 = -25. Do you see why?

eumyang is correct. -52 means - (52) = -25. On the other hand, if you mean the square of -5, write it as (-5)2.
Zula110100100 said:
You could just write [itex] x = \sqrt{25} = \pm 5[/itex]
No, this is WRONG. √25 = 5. Period.

It's true that 25 has two square roots, but the symbol √25 means the principal (or positive) square root. The other square root is written -√25.
 
  • #12
Problem solved. Thank you for all your time in explaining.

I guess I was just a little confused.
 
  • #13
But if we have [itex]\sqrt{x} ; x = 25[/itex] Then the answer is [itex]\pm 5[/itex]? So that it depends on whether we are taking the root of a number or variable? Or is it only if we have a value squared, and we put the squareroot sign that we use [itex]\pm[/itex]?
 
  • #14
Zula110100100 said:
But if we have [itex]\sqrt{x} ; x = 25[/itex] Then the answer is [itex]\pm 5[/itex]?

No. The definition of √x (x > 0) is the positive number y such that y2 = x.

However, if you're given something such as x2 = 100, there exist two solutions. √100, which is +10, and -√100, which is -10
 
Last edited:
  • #15
frozonecom said:
The way I know it, those two are not equal. But my teacher insists on saying that whenever there is a square root, we should always recognize also the negative root.
This is a good question, frozonecom. Apparently there are regional differences, so the correct answer depends on the country you are in (or the country where your teacher studied).

The answers you received in this forum indicate the square root symbol to be taken as denoting the positive square root. I, however, studied along with the same nomenclature as your teacher, and I require recognition of both + and - answers for the explicit square root symbol.

Another area where there are regional differences is in expressions such as 2/5x
On some forums this is staunchly defended as being equal to 0.4/x
On this forum, I'm pleased to see it is just as strongly regarded to be 0.4x

With clarity being crucial to mathematics, I endorse only the latter.
 
  • #16
NascentOxygen said:
This is a good question, frozonecom. Apparently there are regional differences, so the correct answer depends on the country you are in (or the country where your teacher studied).

The answers you received in this forum indicate the square root symbol to be taken as denoting the positive square root. I, however, studied along with the same nomenclature as your teacher, and I require recognition of both + and - answers for the explicit square root symbol.

Another area where there are regional differences is in expressions such as 2/5x
On some forums this is staunchly defended as being equal to 0.4/x
On this forum, I'm pleased to see it is just as strongly regarded to be 0.4x

With clarity being crucial to mathematics, I endorse only the latter.

If clarity is crucial to mathematics, then it would seem that you ought to endorse never typing 2/5x, and always only typing either 2/(5x) or (2/5)x.
 
  • #17
Regarding the original post, if [itex] x - \sqrt{2} = 0 [/itex], then [itex] x = \sqrt{2} [/itex], and that's IT. Surely this is true REGARDLESS of whether you conventionally think of [itex] \sqrt{2} [/itex] as denoting only +1.414..., or whether you think of it as something that could equally well denote either +1.414.. or -1.414...

Either way, the answer is [itex] x = \sqrt{2} [/itex]. An answer of [itex] -\sqrt{2} [/itex] wouldn't make sense, because as dextercioby has already pointed out, plugging this value back into your original equation, you see that it is not a solution:[tex] -\sqrt{2} - \sqrt{2} \neq 0 [/tex]

Now, for those of you who apparently adopt the convention that the radical sign can denote either the positive or the negative root, then I suppose you could argue that the first radical is meant to denote -1.414..., and the second radical is meant to denote +1.414..., in which case you do get 0. However, in that case, you have the same symbol used to represent two different numbers in the same equation, and nothing whatsoever to indicate which one is which. This seems like a really bad notational practice to me.

The convention in which the square root sign denotes only the positive root makes a lot more sense to me.
 
  • #18
Do we do the same thing when it is a trigonometric equation? if:

sin(x) - sqrt(3)/2 = 0

then the solution will be the angles where sin is positive which is on QI and QII.
 
  • #19
frozonecom said:
Do we do the same thing when it is a trigonometric equation? if:

sin(x) - sqrt(3)/2 = 0

then the solution will be the angles where sin is positive which is on QI and QII.

Questions like this also need statement regarding restrictions for x (like -π/2 ≤ x ≤ π/2 or x in ℝ). If there was a restriction of -π/2 ≤ x ≤ π/2, for instance, then you shouldn't look for the angle in Q II at all.
 
  • #20
frozonecom said:
Do we do the same thing when it is a trigonometric equation?
While you are a student, you calculate answers the way your lecturer wants you to. :eek::)

If you are unsure, then I think in an exam it is worth playing safe. Work it out assuming the radical means the positive only. Then, write, "if we are required to also consider the negative of the square root then ..." and provide a further solution.

You are likely to lose few marks if your lecturer doesn't entertain the negative alternative, and you should get full marks if your lecturer expects that you'll examine both outcomes.
 
  • #21
NascentOxygen said:
This is a good question, frozonecom. Apparently there are regional differences, so the correct answer depends on the country you are in (or the country where your teacher studied).

Just out of curiosity, where is it different?
 
  • #22
frozonecom said:

Homework Statement



What I am confused about is:
[itex] x - \sqrt{2}=0[/itex]
[itex]x=\sqrt{2}[/itex]
is it equal to
[itex]x=\pm\sqrt{2}[/itex]

?

Homework Equations


The Attempt at a Solution



The way I know it, those two are not equal. But my teacher insists on saying that whenever there is a square root, we should always recognize also the negative root.

It says based from this post : https://www.physicsforums.com/showpost.php?p=3618911&postcount=2 that it isn't.

Isn't it you only consider to get the positive and negative root if it's like this:
[itex]x^2=25[/itex]
Then that's when you say that [itex]x=\pm5[/itex]

Is there any law or postulate that clears this situation??
Please correct me if I am wrong and if my teacher is actually correct.

And if I am correct, how can I explain it to her so that she'll get it right away? I'm not that good at explaining anyway.

√2 is the length of the hypotenuse of the right angled triangle, where the other two sides are 1 unit long.

None of those values have a positive or negative connotation - they are just values.

if x2 = 2 , then the equation will be satisfied by the positive or negative numbers which are the size of √2, but if we are told x = √2 then it is just a value - like the length of that hypotenuse.
If we wanted it to be the negative value of that size, we would have to say x = -√2

That's the way I see / explain it.

EDIT: we were always told, and taught, if the root sign is already there, it will be stated whether it is positive, negative or both. If you introduce the root sign yourself, you have to allow for both possibilities.

x = √25 means x = 5
x = -√25 means x = -5
x2 = 25 means x = 5 or -5
 
Last edited:
  • #23
PeterO said:
√2 is the length of the hypotenuse of the right angled triangle, where the other two sides are 1 unit long.

None of those values have a positive or negative connotation - they are just values.
I disagree. They are positive values. If we are talking about lengths of things, they are nonnegative numbers.
 
  • #24
e^(i Pi)+1=0 said:
frozonecom,

Only when you take the square root of something do you put ±. If there already is a square root, leave it as it is.

this. I learned this from my teacher.
 
  • #25
Mark44 said:
I disagree. They are positive values. If we are talking about lengths of things, they are nonnegative numbers.

feel free to disagree.

Natural numbers - counting numbers - are neither positive nor negative; there is not even a zero, because if there are none, you can't count them.

We only needed positive and negative when we invented the concept of integers, in order to give closure under subtraction.

As soon as you find a negative length to measure, let me know and I will then see we may need integers for measuring, and not just natural numbers.
 
  • #26
PeterO said:
feel free to disagree.

Natural numbers - counting numbers - are neither positive nor negative; there is not even a zero, because if there are none, you can't count them.

We only needed positive and negative when we invented the concept of integers, in order to give closure under subtraction.

As soon as you find a negative length to measure, let me know and I will then see we may need integers for measuring, and not just natural numbers.

I think what people are asserting here is that natural numbers, being the subset of integers that happen to be positive, are defined to all be positive, and that you cannot have a number of "unspecifiied" or "undefined" sign. I'm inclined to agree with Mark44 on this one, but I'm not a mathematician.
 
  • #27
PeterO said:
feel free to disagree.

Natural numbers - counting numbers - are neither positive nor negative; there is not even a zero, because if there are none, you can't count them.
Not sure how you can say that. Natural numbers form a subset of the set of integers, which consist of positive and negative numbers. Natural numbers are positive.EDIT: Beaten to it by cepheid. :biggrin:
 
  • #28
eumyang said:
Not sure how you can say that. Natural numbers form a subset of the set of integers, which consist of positive and negative numbers. Natural numbers are positive.


EDIT: Beaten to it by cepheid. :biggrin:

they say great minds think alike...

...and fools seldom differ.:rofl:
 
  • #29
PeterO said:
Natural numbers - counting numbers - are neither positive nor negative; there is not even a zero, because if there are none, you can't count them.

Ignoring the 'neither positive nor negative' comment, this still is not necessarily true. It is fairly common to define N to include 0. I know this is completely standard in mathematical logic and I have seen many mathematicians in other fields use this convention as well.
 
  • #30
jgens said:
It is fairly common to define N to include 0.
While I am aware of this definition, in the school math textbooks that I have seen, N does not include 0. However, there is another set, whole numbers, which consists of the set of natural numbers and 0.
 
  • #31
That is interesting. I have never seen the term 'whole numbers' in any real math textbook. Also, I think the convention for taking N to include 0 is more recent, so that might account for some of the difference.

On an unrelated note, the logicians I have met start counting from zero. That is, textbooks start with chapter 0. When they list sequences they being at x0. When they number properties, they start with 'property 0' o_O
 
  • #32
PeterO said:
feel free to disagree.

Natural numbers - counting numbers - are neither positive nor negative;
This is nonsense. This would imply that, for example, 5 cannot be placed anywhere on the real number line.
PeterO said:
there is not even a zero, because if there are none, you can't count them.

We only needed positive and negative when we invented the concept of integers, in order to give closure under subtraction.
PeterO said:
As soon as you find a negative length to measure, let me know and I will then see we may need integers for measuring, and not just natural numbers.
I already touched on this in post 23.
Mark44 said:
If we are talking about lengths of things, they are nonnegative numbers.
 
  • #33
Mark44 said:
This is nonsense. This would imply that, for example, 5 cannot be placed anywhere on the real number line.

I always thought Real numbers - another conceptual imagination of man to give closure under involution - had a positive and negative sub-set, so of course will included the integer +5, not the Natural Number 5.
 
  • #34
PeterO said:
I always thought Real numbers - another conceptual imagination of man to give closure under involution - had a positive and negative sub-set, so of course will included the integer +5, not the Natural Number 5.

1) What does involution have to do with R? The special thing about R is that it is the unique ordered field with the least upper bound property.

2) The notion of signless natural numbers is not standard. There are several reasons for this. First, we like to view the natural numbers as a subset of Z, Q, R, etc. Second, one way of defining an ordered field involves specifying a collection of positive numbers that satisfy certain properties. The standard order on N gives the natural numbers all of these properties, and so it makes sense to call them positive.
 
  • #35
PeterO said:
I always thought Real numbers - another conceptual imagination of man to give closure under involution - had a positive and negative sub-set, so of course will included the integer +5, not the Natural Number 5.
The integer +5 (the plus sign is unnecessary) and the natural number 5 can both be found at exactly the same place on the real number line.

I have no idea what "closure under involution" means. This didn't come up in any of the numerous math classes I took.
 

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