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manjuvenamma

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First of all, my sincere apologies. As I do not know how to post a new problem. I am not anywhere seeing 'POST YOUR QUESTION' though I am able to see VIEW YOUR POSTS.

My question is to find the value of sqrt ( 6 + sqrt (6 + sqrt (6 + sqrt (6 + ....... ).
I solved it assuming it x. then we get a quadratic equation square(x) = 6 + x. By solving this quadratic equation we get two values -2 and 3. We say 3 is the value as it is positive.
But my dilemma is how come we ended up with -2 as a possible solution. Every square root of a positive rational number has two values +/- (some rational number). Here in stead of getting +3 and -3 which should be the square root (6 + sqrt (6 + sqrt(6 + ..... we ended up with -2 and 3. While 3 looks reasonable, where is the logical flaw that causes -2 to appear as one of the 2 values. Thanks and also please tell me how to post a new question.

sqrt stands for square root.
 
Equation: x^2 - x - 6 = 0 ; x = 3 or x = -2

Substitute x = -2 in equation:
(-2)^2 -(-2) - 6 = 0
4 + 2 - 6 = 0
0 = 0

Now do you see why it's a valid solution? :idea:
 
manjuvenamma said:
First of all, my sincere apologies. As I do not know how to post a new problem. I am not anywhere seeing 'POST YOUR QUESTION' though I am able to see VIEW YOUR POSTS.

My question is to find the value of sqrt ( 6 + sqrt (6 + sqrt (6 + sqrt (6 + ....... ).
I solved it assuming it x. then we get a quadratic equation square(x) = 6 + x. By solving this quadratic equation we get two values -2 and 3. We say 3 is the value as it is positive.
But my dilemma is how come we ended up with -2 as a possible solution. Every square root of a positive rational number has two values +/- (some rational number). Here in stead of getting +3 and -3 which should be the square root (6 + sqrt (6 + sqrt(6 + ..... we ended up with -2 and 3. While 3 looks reasonable, where is the logical flaw that causes -2 to appear as one of the 2 values. Thanks and also please tell me how to post a new question.

sqrt stands for square root.
Click to expand...

----------------

I don't believe there is any logical flaw at all. Technically, the value -2 is *also* a legitimate solution *unless* the sqrt function is meant to produce only the principal (positive) square root of its argument.

In my experience, "sqrt" is indeed a single value function producing only a positive value. Therefore you could reject the -2 value as an impossible value for "sqrt" to produce.
 
Yes, -2 also satisfies the quadratic equation we obtained.

But, what I am referring to as logical flaw is this :

Square root of any real number is +/- (another real number). They are opposite in sign but equal in absolute value.
For example, square (x) = 16 has two solutions +/- 4.
square of (x) = 8 has two solutions, +/- 2 * square root(2)
6+sqrt(6+ sqrt(6+....) is also a real number.
Then how come, its square roots are +3 and -2 in stead of +3 and -3 or +2 and -2.

Sorry, but none of the replies address this question I raised.
 
manjuvenamma:

In my first reply, I argued that there is no "logical flaw" if we interpret sqrt() to be a single-valued function that delivers only the "principal" (non-negative) root of its argument.

On the other hand, if we insist, as you do, that sqrt() is properly a 2-valued function, delivering both a negative and positive root, then there is *also* no logical flaw. In that case, +3, -3, +2, and -2 are all valid solutions for x, where x = sqrt(6+sqrt(6+sqrt(6+ ...

Here's why:

Let's take +2 or -2 as *both* being solutions for x. Then let's test for equality in the expression:

+2 =? sqrt(6+sqrt(6+sqrt(6+

Begin by squaring both sides. Then we get

+4 =? 6+ [sqrt(6+sqrt(6+...]

Now, for the expression in brackets, we can substitute *either* +2 or -2 since both are legitimate. If we choose to substitute -2, we do indeed get an identity, which is what we sought to do. So +2 is a valid solution.

The same conclusion works for -2, +3, and -3 as solutions for x so long as we are free to choose either the positive or negative value of sqrt(). So if we're consistent in using sqrt() as a two-valued fucntion, there is no logical flaw.

Does this make sense?
 
Dr.Lee,

Thank you for your time and interest.

We started with wanting to find the square root of a real number : 6 + sqrt(6 +.........). (Or is it not a real number, really?) We ended up with two values -2 and +3 and (may be their sign opposites if you want). The answers satisfy the quadratic equation but that does not mean that they are right. If 3 is taken as the answer, then 6+... (what was inside teh square root should be 9. If -2 is taken, then it should be 4. What is under the square root, how can it be both 9 and 4.

If we come back and see where we started. Can the square roots of a real number be two real numbers that are different in magnitude?

Also no one has answered to my question 'how to post a new question?' So I am continuing this chain for a new question.

We all know how to convert a terminating or recurring decimal into a rational number. The method gives correct results for say 0.11111 = 1/9. But apply the method for 0.999999 and it gives you the wrong answer saying it is equal to 1. Why did it fail?

Best regards.
 
manjuvenamma said:
Dr.Lee,

Thank you for your time and interest.

We started with wanting to find the square root of a real number : 6 + sqrt(6 +.........). (Or is it not a real number, really?) We ended up with two values -2 and +3 and (may be their sign opposites if you want). The answers satisfy the quadratic equation but that does not mean that they are right. If 3 is taken as the answer, then 6+... (what was inside teh square root should be 9. If -2 is taken, then it should be 4. What is under the square root, how can it be both 9 and 4.

If we come back and see where we started. Can the square roots of a real number be two real numbers that are different in magnitude?

Also no one has answered to my question 'how to post a new question?' So I am continuing this chain for a new question.

We all know how to convert a terminating or recurring decimal into a rational number. The method gives correct results for say 0.11111 = 1/9. But apply the method for 0.999999 and it gives you the wrong answer saying it is equal to 1. Why did it fail?

Best regards.
Click to expand...

Well, you ask some interesting questions.

First, the expression x = sqrt(6+sqrt(6+ ... does indeed represent a real number *if* we stipulate that sqrt() is a single valued function that yields either a single non-negative number. Once we require that sqrt() be a two-valued function, then at every point where we compute a square root, we spawn two possible values to be plugged into the next term in the infinite expansion. You cannot simultaneously require that 'x' be a single number *and* require that sqrt() be a two-valued function.

As far as the apparent paradox on .99999... , actually the method does not fail. The value of .99999... is not equal to 1 if you terminate the series early. The "...." is meant to represent an infinite sequence. In the limit, as the number of 9's approaches infinity, the value of the expression approaches 1 to within an arbitrarily small number (usually called epsilon). Similarly, .111111... does not represent 1/9 if you terminate the series early. However, it approaches 1/9 in the limit as the number of 1's approaches infinity.

These look like pre-calculus questions because the concept of 'limit' keeps coming up. Is that what you're studying?

These are good questions you ask ... I'm glad to see that you are seeking real understanding ... to me, this shows a strong math aptitude!

I'm newly retired after 40 years in engineering, and I'm also new to this forum ... the best thing to do to find the variations on question-posting is to send an email to the one of the principals who run the forum ...

Regards,

Dr Lee
 
mmm4444bot,

I did not know how to start a new question separately. I tagged on to some other question and some one separated it as a new question and poste it. In the last post too, I started a new question along with a reply to an older question.

Hope you are clear now.

Dr. Lee,

Thanks for your patience and understanding.

I am a post-graduate in microwave and radar engineering and am working for the last about 25 years in R & D. About 10 more years are left in me before I retire. I love logic and problem solving. Hence, I love the 'queen of the sciences' which is based on deductive reasoning. Though, my job does not require it, I take interest in math and revisit math and enjoy its beauty.

Coming back to our question, I also thought of "sticking to one-valued or two-valued throughout'. So indeed, if we are allowing two values, then x we are talking about is not a real number, but it is several real numbers in "disguise" as it depends on which of the two values we choose to substitute at all the infinite square roots that occur in x.

Once we accept x as not a one real number, then can we tell on which substitutions it becomes equal to 2? Clearly, when we take sqrt(6) = 2.4494..., it becomes 3.

About the other question on rational numbers, 0.1111.... comes indeed if we divide 1 by 9 (long division method) or for that matter any rational number you get in this method. But 0.9999.... is not obtained from 1 (by long divison). May be in this case we can not find two natural numbers p and q such that 0.999.... is obtained by long dividing p with q. I agree with your statements on 'limit' and convergence.

Best regards,
 
Denis,

I got it now.
I mistook it for a new topic/section and I was searching for 'POST your question' like 'VIEW YOUR POST'
Thanks for the help.

Best regards.
 
manjuvenamma said:
… Hope [it is] clear [to you] now …
Click to expand...


Quite clear; thank you.

(I regret that the moderator who reposted your original did not insert "SPLIT" into the subject line, as required by "law".) 8-)

 
Dimension

Hi

Will you give me the dimension (length and width) of a rectangle which has the area and perimeter.
The Area = 12 and Perimeter = 14 Will you also draw the rectangle?
 
Re: Dimension

wrb said:
Hi

Will you give me the dimension (length and width) of a rectangle which has the area and perimeter.
The Area = 12 and Perimeter = 14 Will you also draw the rectangle?
Click to expand...

Do you understand my question
 
Mountainside fleece sold 40 neckwarmers. Solid neckwarmers sod for $9.90 and print sold for $12.75 each. In all, $421.65 was taken in for the neckwarmers. How many of each type is sold. How do I work this problem?
 
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