why is #A wrong?

allegansveritatem

Full Member
Joined
Jan 10, 2018
Messages
962
Here is the multiple choice problem:
mistake2.jpg

I chose A as being correct but worked out the others to be sure and found that B is the correct one or at least seems to have a little more claim to that status in that when I checked A on my calculator it said True for 7i but false when I tried -7i. Here is what I get when I work out A:

mistake.jpg

What is wrong with this?
 
Here is the multiple choice problem:
View attachment 10989

I chose A as being correct but worked out the others to be sure and found that B is the correct one or at least seems to have a little more claim to that status in that when I checked A on my calculator it said True for 7i but false when I tried -7i. Here is what I get when I work out A:

View attachment 10990

What is wrong with this?


\(\displaystyle \sqrt{49} = 7\) No plus-or-minus about it. It is a common misunderstanding.
 
Here is the multiple choice problem:
View attachment 10989

I chose A as being correct but worked out the others to be sure and found that B is the correct one or at least seems to have a little more claim to that status in that when I checked A on my calculator it said True for 7i but false when I tried -7i. Here is what I get when I work out A:

View attachment 10990

What is wrong with this?
I will give my answer with which some will take issue (i.e.@lookagain).
I worked with and taught foundation of mathematics for both in mathematics & philosophy departments.
So my take: \(\displaystyle \pm 2\) is a symbol, a short hand notation, for "positive two or negative two" \(\displaystyle 2 \vee -2\).
So the short cut \(\displaystyle \sqrt{16}=\pm 4\). But in my advanced courses I would mark that incorrect, zero credit.
\(\displaystyle \sqrt{9}\) is a number, one number not two numbers. The square roots of \(\displaystyle 9\) are \(\displaystyle \pm 3\) .
Note how the verb is plural because there are two of them. But \(\displaystyle \large\sqrt 9=3\) and \(\displaystyle \large\sqrt 9\ne -3\).

To an answer for your question.
1) I taught that the radical for even roots applied only to non-negative real numbers and gives one number.
Thus \(\displaystyle \sqrt 9=3\) and \(\displaystyle \sqrt[3]{-27}=-3\)

2) There are two square roots of \(\displaystyle -49\) which are \(\displaystyle \pm 7\bf{i}\) but neither is \(\displaystyle \sqrt {-49}\) because that would be a notation error.

3) There are in the complex field there are \(\displaystyle n\text{ nth roots of any number}\)
So what are the cube roots of \(\displaystyle -27~?\). Well SEE HERE. There are three.

4) There are six sixth roots of \(\displaystyle -64\) See Here Note that \(\displaystyle 2\bf{i}\) is one of the roots as well as \(\displaystyle -2\bf{i}\).
But it would be a mistake to use the notation \(\displaystyle \sqrt[6]{-64}=\pm 2\bf{i}\). Do you see why?

 
doesn't (-7)2​=49?
Read carefully my reply.
Yes \(\displaystyle (-7)^2=49\) but \(\displaystyle \sqrt{49}\ne -7\) because the latter is a notation error.
 
doesn't (-7)2​=49?

Read pka VERY CAREFULLY. Your question is not actually relevant.

The problem is \(\displaystyle \sqrt{49} = ??\)? NOT, Find solutions to \(\displaystyle x^{2} = 49\)
 
Here is the multiple choice problem:
View attachment 10989

I chose A as being correct but worked out the others to be sure and found that B is the correct one or at least seems to have a little more claim to that status in that when I checked A on my calculator it said True for 7i but false when I tried -7i. Here is what I get when I work out A:

View attachment 10990

What is wrong with this?

B is definitely correct; the issue is whether A is also correct.

What you wrote is wrong because the radical \(\displaystyle \sqrt{49}\) represents only one value (making it a function), not both square roots. It's not clear what you actually did with your calculator.

I agree in general with what pka said: At an advanced level, A can be considered (sort of) correct, in that there is no single value that can be called THE square root of a negative number (or, more generally, of a complex number), so the best you can do is to list all roots. But it is not really proper at your level to use the radical to mean this. The proper thing to do is to say that the square roots of -49 are 7i and -7i.

Unfortunately, many textbooks at the level of high school or college algebra try to simplify things by saying that \(\displaystyle \sqrt{-1} = i\), and \(\displaystyle \sqrt{-a^2} = ai\), as if this were the positive (principal) root as it is when working only with real numbers. Imaginary numbers can't be positive! But pretending it is, allows them to make the radical a function, and avoid having to talk about "multiple-valued functions".

Presumably your textbook is using this convention, so that they consider A to be false because \(\displaystyle \sqrt{-49} = 7i\) only. Or, they may be taking pka's convention (as I understand it) and just saying that you can't write such a radical at all. (That, unfortunately, makes it impossible to write the quadratic formula, if you take it seriously.)

You'll just have to go along with what your textbook says until you get into a Complex Variables course, when they may allow you to call A correct.
 
that they consider A to be false because \(\displaystyle \sqrt{-49} = 7i\) only. Or, they may be taking pka's convention (as I understand it) and just saying that you can't write such a radical at all. (That, unfortunately, makes it impossible to write the quadratic formula, if you take it seriously.)
You'll just have to go along with what your textbook says until you get into a Complex Variables course, when they may allow you to call A correct.
Although I have been retired from active research since 2005, publishers still send me books to review.
When I read a highschool text, I pay attention to notation. If the author says \(\displaystyle \sqrt{-49}=7\bf{i}\), I asks the author why not say \(\displaystyle \sqrt{-49}=-7\bf{i}\) after all \(\displaystyle (-7\bf{i})^2=-49\) RIGHT? So doing my part to correct notation.

Unfortunately more often than not I get very negative "blow-back" on that comment.
Here is a topical one: we all know that\(\displaystyle \sqrt{x^2}=\pm|x|\) , that has always been the case. Has it not?
For years we have taught that \(\displaystyle \sqrt{49}=\pm 7\), why are you asking us to change now?
 
Here is how I handle such problems. I use the fact that if a*b are NOT both negative, then sqrt(a*b) = sqrt(a)*sqrt(b) as well as the fact that sqrt(-1) = i.
So sqrt(-49) = sqrt(49*(-1))= sqrt(49)*sqrt(-1) = 7i

Now for many years I thought that was correct until I joined this forum and saw a problem like this: sqrt(25/-1) = sqrt(25)/sqrt(-1) = 5/i = -5i. Or is it sqrt(25/-1) = sqrt(-25) = sqrt(25)*sqrt(-1) = 5i. I am getting different answers.

So in the above, we can't have a negative in the denominator.
 
B is definitely correct; the issue is whether A is also correct.

What you wrote is wrong because the radical \(\displaystyle \sqrt{49}\) represents only one value (making it a function), not both square roots. It's not clear what you actually did with your calculator.

I agree in general with what pka said: At an advanced level, A can be considered (sort of) correct, in that there is no single value that can be called THE square root of a negative number (or, more generally, of a complex number), so the best you can do is to list all roots. But it is not really proper at your level to use the radical to mean this. The proper thing to do is to say that the square roots of -49 are 7i and -7i.

Unfortunately, many textbooks at the level of high school or college algebra try to simplify things by saying that \(\displaystyle \sqrt{-1} = i\), and \(\displaystyle \sqrt{-a^2} = ai\), as if this were the positive (principal) root as it is when working only with real numbers. Imaginary numbers can't be positive! But pretending it is, allows them to make the radical a function, and avoid having to talk about "multiple-valued functions".

Presumably your textbook is using this convention, so that they consider A to be false because \(\displaystyle \sqrt{-49} = 7i\) only. Or, they may be taking pka's convention (as I understand it) and just saying that you can't write such a radical at all. (That, unfortunately, makes it impossible to write the quadratic formula, if you take it seriously.)

You'll just have to go along with what your textbook says until you get into a Complex Variables course, when they may allow you to call A correct.

I think you are saying that it is a matter of convention, at least as far as a novice goes, and I will accept that. If my author likes it, s'OK with me.
 
I will give my answer with which some will take issue (i.e.@lookagain).
I worked with and taught foundation of mathematics for both in mathematics & philosophy departments.
So my take: \(\displaystyle \pm 2\) is a symbol, a short hand notation, for "positive two or negative two" \(\displaystyle 2 \vee -2\).
So the short cut \(\displaystyle \sqrt{16}=\pm 4\). But in my advanced courses I would mark that incorrect, zero credit.
\(\displaystyle \sqrt{9}\) is a number, one number not two numbers. The square roots of \(\displaystyle 9\) are \(\displaystyle \pm 3\) .
Note how the verb is plural because there are two of them. But \(\displaystyle \large\sqrt 9=3\) and \(\displaystyle \large\sqrt 9\ne -3\).

To an answer for your question.
1) I taught that the radical for even roots applied only to non-negative real numbers and gives one number.
Thus \(\displaystyle \sqrt 9=3\) and \(\displaystyle \sqrt[3]{-27}=-3\)

2) There are two square roots of \(\displaystyle -49\) which are \(\displaystyle \pm 7\bf{i}\) but neither is \(\displaystyle \sqrt {-49}\) because that would be a notation error.

3) There are in the complex field there are \(\displaystyle n\text{ nth roots of any number}\)
So what are the cube roots of \(\displaystyle -27~?\). Well SEE HERE. There are three.

4) There are six sixth roots of \(\displaystyle -64\) See Here Note that \(\displaystyle 2\bf{i}\) is one of the roots as well as \(\displaystyle -2\bf{i}\).
But it would be a mistake to use the notation \(\displaystyle \sqrt[6]{-64}=\pm 2\bf{i}\). Do you see why?


I catch your drift but only now and then. I am on rung two of a ladder that has more rungs than I can count.
 
Thanks to all who contributed to this thread. A lot of what was said was over my head but I enjoy such discussions as giving me some idea how mathematicians think. I haven't got enough time left this time around to accomplish it, but if I have to be born again (Heaven Forbid!) I would be happy to spend my next life studying this subject. Of course, even a whole lifetime spent studying it would not be enough to do it justice.
 
Reading over these posts it occurs to me that the gist of what I'm being told is this: I have been treating this expression as though it had been forged in the fiery furnace of a quadratic equation when it is actually just the representation of a single entity. It represents itself, not the output of a process....Well, something like that.
 
Reading over these posts it occurs to me that the gist of what I'm being told is this: I have been treating this expression as though it had been forged in the fiery furnace of a quadratic equation when it is actually just the representation of a single entity. It represents itself, not the output of a process....Well, something like that.

Something like that ...

I'd say that the radical is a notation that has been given to you, which, by convention, refers only to one value (as defined in your textbook); if it had been something you yourself created in the process of solving an equation, you would have had to include the plus-or-minus.
 
Top