Real Number Operations Caveat

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Master_Pedant
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13 Dec 2011, 9:17 pm

Okay, I've been reading this website:

http://www.vitutor.com/arithmetic/real_ ... mbers.html

Which says:

Quote:

With real numbers all operations can be performed, except for the root of an even index and negative radicand, and division by zero.

What does "root an even index and negative radicand" exactly mean?

For instance, would this be an "even index and negative radicand"

√-2

And, if so, would this be trying to "root and even index and negative radicand"?

√√-2

Or am I hopelessly confused?

Upon doing some reading, I know that an "index" apparently is the exponent of 1/2 to which -2 is raised (represented via the root symbol), so I guess that really wouldn't be "an even index" (according to this). What, then, would be an example of an "even index and negative radicand"?

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dragonbean
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13 Dec 2011, 11:50 pm

An example of a "root of an even index and negative radicand" is √-2 .
Because the radicand, the 2, is negative, and because you're taking the square root (meaning the root is of an even index).

You can take the cubic root of a negative index. For example, the cubic root of -64 is -4. In this case, the root is of an odd index (3). You can't take the square root of -64, because the root is an even index (2).

Did that clarify things or mess them up more?

And you're right, that website did have misleading wording with that sentence.

Last edited by dragonbean on 14 Dec 2011, 8:59 pm, edited 1 time in total.

Master_Pedant
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14 Dec 2011, 12:33 am

dragonbean wrote:

An example of a "root of an even index and negative radicand" is √-2 .
Because the radicand, the 2, is negative, and because you're taking the square root (meaning the root is of an even index).

You can take the cubic root of a negative index. For example, the cubic root of -64 is 4. In this case, the root is of an odd index (3). You can't take the square root of -64, because the root is an even index (2).

Did that clarify things or mess them up more?

And you're right, that website did have misleading wording with that sentence.

That clarified things, matters!

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ruveyn
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14 Dec 2011, 7:53 am

A 2 * k th root where k is an integer >= 1.

Roots odd indices exist for all real numbers. Do you know why? (I leave that for you as an exercise).

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dragonbean
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14 Dec 2011, 9:17 pm

Quote:

That clarified things, matters!

That's good ^^ Also i messed up typing, the cubic root of -64 is -4, not 4, but i edited that.

Quote:

Roots odd indices exist for all real numbers. Do you know why? (I leave that for you as an exercise).

I'm not sure if I understand your question and I'm not sure if you're asking me or Pedant, but I'm bored, so: because a negative times a negative is a positive, and a negative times a positive is a negative. The radicand can be negative and at the same time have a real number as its root because the negative real root is multiplied by itself an odd number of times if the root index is odd, resulting in that negative radicand.

jackmt
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24 Dec 2011, 3:11 am

Master_Pedant wrote:

Okay, I've been reading this website:

http://www.vitutor.com/arithmetic/real_ ... mbers.html

Which says:

Quote:

With real numbers all operations can be performed, except for the root of an even index and negative radicand, and division by zero.

Concerning division by 0; I wrote a linguistic proof that 0^0 (zero raised to the zero power) = 1 and is therefore a legitimate divisor. Any one agree? disagree?

nat4200
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24 Dec 2011, 5:04 am

Redacted

Last edited by nat4200 on 19 Apr 2012, 5:51 am, edited 1 time in total.

ruveyn
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24 Dec 2011, 12:03 pm

Master_Pedant wrote:

Okay, I've been reading this website:

http://www.vitutor.com/arithmetic/real_ ... mbers.html

Which says:

Quote:

With real numbers all operations can be performed, except for the root of an even index and negative radicand, and division by zero.

e. g. 10 th root of -1.

The basic reason is that all even powers of negative numbers are positive. That is why the real number system has to be extended to include numbers who even powers are negative.

ruveyn

jackmt
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24 Dec 2011, 2:09 pm

nat4200 wrote:

jackmt wrote:

Master_Pedant wrote:

Okay, I've been reading this website:

http://www.vitutor.com/arithmetic/real_ ... mbers.html

Which says:

Quote:

With real numbers all operations can be performed, except for the root of an even index and negative radicand, and division by zero.

Concerning division by 0; I wrote a linguistic proof that 0^0 (zero raised to the zero power) = 1 and is therefore a legitimate divisor. Any one agree? disagree?

@jackmt
What number, in the Real Number System (this being the number system being discussed), do you suggest should be the answer for: 7 divided by 0? And why?
NB: If your answer should involve a value of infinity, then please understand that infinity as a number is not a value defined in the set of "real" numbers.

Division by zero is undefined. Agreed? I asserted that zero to the zero power is 1, and therefore allowed as a divisor. The exponent y in x^y is typically understood as declaring how many times x is used in the expression as a factor. This is not quite right, but let's accept it. 0^0 says that zero is used zero times. How is it then division by zero? Why is it the one exception to the paradigm when there is (it seems to me) no good reason to exclude it? What problems does it create? I have talked to many mathematicians. A few of them agree. Most of those who disagree seem to have a visceral and angry reaction (inappropriate for a strictly cerebral discipline) and many just fail to understand what I am asserting. One, after his visceral reaction, acknowledged that in some systems it is treated as one. And within several systems it acts as one in limited applications.

ruveyn
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24 Dec 2011, 3:02 pm

jackmt wrote:

This is not quite right, but let's accept it. 0^0 says that zero is used zero times. How is it then division by zero?

0^0 = 0^1 / 0^1 = 0 / 0 = undefined. If you want to keep the law of exponents don't do 0 ^ 0.

By the way - a single contradictions destroys the entire system. Consistency is an absolute requirement.

ruveynb

jackmt
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24 Dec 2011, 7:50 pm

ruveyn wrote:

jackmt wrote:

This is not quite right, but let's accept it. 0^0 says that zero is used zero times. How is it then division by zero?

0^0 = 0^1 / 0^1 = 0 / 0 = undefined.
ruveynb

This equation is illegitimate for at least 3 reasons. First, on this model all 0^n is undefined as it may be represented as

0^n = 0^((n+1) -1) = 0^(n+1)/0^1

And we don't want to say that all 0^n is undefined, but that it is 0 for all n not equal to zero. So 0^0 is already an exception to a rule. My claim is that it an exception to a different rule; namely, that it is the only value of 0^n that may be used as a divisor.

Second, it defines the simple in terms of the complex and in terms of itself, as if we were to define 1 as:

1 = 1-(1-1).

True enough, but what have we gained?

Third, it comes dangerously close to, if not actually, committing the fallacy of petitio principii: division by zero is already proscribed.
There are other problems as well.

The problem most mathematicians seem to have with my claim is the feeling that the outward expression, being composed of only zeroes, must represent zero underlyingly somehow. But a positive number may be expressed as a double negative.

There is much more to say on the matter. Where does the '1' come from in x^0=1, for instance? The standard answer to this that it behaves as 1. This is weak. It behaves as 1 because it is 1.

In my original claim I said I have a linguistic proof; i.e., 0^0=1 for grammatical reasons. It completely neatens the paradigm, leaving only the one above mentioned exception.

ruveyn
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25 Dec 2011, 3:10 am

jackmt wrote:

ruveyn wrote:

jackmt wrote:

This is not quite right, but let's accept it. 0^0 says that zero is used zero times. How is it then division by zero?

0^0 = 0^1 / 0^1 = 0 / 0 = undefined.
ruveynb

In mathematics linguistic proofs are dreck.

What you have presented is mostly nonsense.

For example 0^1 = 0, In fact for n > 0 0^n = 0. Why? 0*0*...*0 (n terms) = 0.

ruveyn

jackmt
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25 Dec 2011, 1:08 pm

ruveyn wrote:

jackmt wrote:

ruveyn wrote:

jackmt wrote:

This is not quite right, but let's accept it. 0^0 says that zero is used zero times. How is it then division by zero?

0^0 = 0^1 / 0^1 = 0 / 0 = undefined.
ruveynb

In mathematics linguistic proofs are dreck.

What you have presented is mostly nonsense.

For example 0^1 = 0, In fact for n > 0 0^n = 0. Why? 0*0*...*0 (n terms) = 0.

ruveyn

Math is abstracted from language. It is a language with its own syntax, logical operators (verbs), substantives (e.g. numbers), pronouns, (variables), negation, etc. It also has implicit, i.e., non-overt elements: 1 for example, is deleted as multiplier because it is the identity element of multilplication. Similarly with zero and addition. This is not dreck.

And I know that 0^n = 0. You answer as if I had made some simple grade school math error, but you do not even address what I have said. Please check your attitude and look again. And what is nonsensical about what I presented? If it is wrong, show me. But don't insult me and dismiss me. It makes me think you can't answer me. You haven't so far.

lau
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25 Dec 2011, 3:24 pm

jackmt wrote:

ruveyn wrote:

...

None of what you say has any weight.

"0^n" is zero, provided "n" is not zero.

With "z" in the complex domain, even...

"0^z" is zero, provided "z" is not zero.

You have made a "simple grade school math error".

"z^0" is unity, provided "z" is not zero.

0^0 is undefined.

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jackmt
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25 Dec 2011, 8:39 pm

lau wrote:

jackmt wrote:

ruveyn wrote:

...

You cannot seem to think beyond what you have been taught. If you disagreed that would be one thing, but clearly you cannot comprehend; in your 3 responses you have not even once addressed what I have said. I'm done.

ruveyn
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25 Dec 2011, 9:52 pm

jackmt wrote:

What you have said makes no mathematical sense.

ruveyn

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