Complex no

[JeffM]

why does -3^2 translate to -9 .
I thought -3^2 meant (-3)^2 which gives me 9 . Is there any connection of Bedmas with the evaluation of this expression.

First, BEDMAS is a simplifying acronym designed to help beginning students.

Second, the minus sign means has at least three different meanings in mathematics. When it is placed BETWEEN TWO number indicators, it means to do the operation of subtraction. It is being used as a BINARY operator which works on two numbers. When it is placed IN FRONT OF A SINGLE number indicator, it means to do the operation of negation. When it is placed in front of a numeral, it also indicates that the number is less than zero; it is being used as a relational symbol like = or < as well as an operator.

In the order of operations, negation comes after multiplication and division but before addition and subtraction.

So [MATH]- 3^2 * 2 = - (3^2) = - (9) = - 9 \ne (- 3)^2.[/MATH]
This means that [MATH- 3^2 + 20 = 11 = 20 - 3^2.[/MATH]

Suppose A is my proposition "c and b are positive real no's and X is a real no" and B is another proposition "the identities are valid "

Then If A is true I see then B has to be true for any combination of no's I take; with keeping in mind the values are taken matching with the constraint in proposition A.

But there is no A =true B= False Case according to my experiment .

Now if A is false then B can be true still.
I checked.(this was the misconception.i had)

And another case can be both A and B are false which they showed to prove the failure of the law.


Ok the law is valid for all Integer exponent and b and c must be non zero Complex no's .

If you are prime minister of India, then you an Indian citizen.

You are not prime minister of India.

Therefore, you are not an Indian citizen.

It is a logical fallacy that

A entails B
A is false
Thus, B is false.

EDIT: Dr. Peterson and I agree. He says to treat negation as though it were multiplication by negative 1. I say it comes beteen multiplication and division. The results are the same.

 

[Saumyojit]

In the order of operations, negation comes after multiplication and division but before addition and subtraction

Dr p has written Bendmas which means negation comes before multiplication and division.
I never learned this .Wow.

So −32∗2=

2 is typo I think!

another law I just came across.

a^b=a^c =>b=c (a>0 and a>1)

Can 'a' be negative real no .
Or it has to be positive only.

 

[Dr.Peterson]

Dr p has written Bendmas which means negation comes before multiplication and division.
I never learned this .Wow.

Absolutely not!

Do you think BEDMAS means that division comes before multiplication? It does not.

Considering that, what do you think I really meant? (And why do you think I despise things like BEDMAS?)

a^b=a^c =>b=c (a>0 and a>1)

Can 'a' be negative real no .
Or it has to be positive only.

As always, this depends on context, and in particular on what has been said about the variables.

If b and c can be any real number, then there are problems defining a^b and a^c uniquely. That's one reason they'd omit negative bases. (Another is that you'd want to use logarithms to prove it. What would go wrong then?)

If b and c are integers, then I'll let you think about it. Can we relax the restriction on a and let it be negative?

Can you show us where you found this one, so we can check out what else was said?

 

[JeffM]

@Saumyojit

I must apologize for my post # 21. I was in the middle of proofreading it when my wife shoved my baby grandson in my lap to watch for an hour.

First, yes, the * 2 was a typo. Furthermore, there was a more important typo. I meant to say that you can think of negation as coming after multiplication and division but before addition and subtraction (as I tend to do) or as part of multiplication and division by equating negation to multiplication by minus 1 as Dr. Peterson suggested. Either way works. In any case, exponentiation precedes it.

 

[Dr.Peterson]

I'm reminded of this answer I wrote long ago: Precedence of Unary Operators. As JeffM said, both ways of seeing negation give the same result, which fortunately agrees with other reasons for this choice.

On the other hand, it is so easy to misread an expression like -3^2 that some textbooks I've seen avoid such expressions entirely, and others just recommend always using parentheses.

 

[Saumyojit]

Do you think BEDMAS means that division comes before multiplication? It does not.

Considering that, what do you think I really meant? (And why do you think I despise things like

I don't know why u despise
But I got the point of negation and it's precedence

If b and c can be any real number, then there are problems defining a^b and a^c uniquely

In the book it is clearly given

When base a is any real no and exponent b and c are Integrs under this they stated 6 laws of Exponention

Now my objective is to find a counter example with base any real no and exponent b and c non integr .
I tried using this eg
(-4)^1/2 =(-4)^1/2 but it works.
(-8)^1/3=(-8)^1/3 but it works
Can u give me one counter example where exponent are non integer with any real base
and the law gets invalid.

Why did u said " If b and c can be any real number, then there are problems defining a^b and a^c uniquely" ?

Are u saying that If a=4 and b&c is 1/2 ;

4^(1/2)=4^(1/2)
Then on one side 2 may come and on the right side -2 will come leading to ambiguity.

But my reasoning is wrong as you told "they would omit negative base" but 4 is positive base .
then please give a Eg to understand.

. (Another is that you'd want to use logarithms to prove it. What would go wrong then?)

I will be starting log chapter but what are you saying here . I can use log laws to prove the identity law is invalid or valid ??

 

[JeffM]

I cannot know why Dr. Peterson despises BEDMAS, but I suspect it is because it is a easily memorizable simplification that teaches a mechanical rule without explaining anything. He wants people to understand, not parrot.

Exponents were first defined in terms of positive integers greater than one applied to rational numbers; it was defined as repeated multiplication of the base. Combinatorics and the laws of exponents themselves quickly suggested the urility of defining exponents in a different way, a way that preserved the laws of exponents and the original meaning when it applied, but covered all rational numbers.

It was not until the late 19th century that real numbers were fully understood; they correspond to no observable physical facts; they are idealized abstractions of the human mind. Unless you have studied some form of real analysis, it is impossible to prove anything about irrational numbers raised to an irrational power.

I have no idea why the author of your book limited that proposition to integers. Probably because it is easily verifiable by numerical experiment. As was explained before, asserting that something is true in one domain does NOT constitute an assertion that it is false in any other domain.

 

[Dr.Peterson]

I don't know why u despise
But I got the point of negation and it's precedence

I intentionally overstated my opinion using the word "despise"; but I have seen too many students who think it is the entire rule, and therefore think that BEDMAS means D before M (or that PEMDAS means M before D!), when it is intended just as a reminder of a way of seeing expressions that is significantly more nuanced than six letters can fully express. I always teach the concept first, and then tell students who have heard of PEMDAS that they may use it as long as they know the real facts it represents. (But when I see students write "PEMDAS" at the top of their exam paper to remind them of the rules, it makes me worry that they haven't really internalized it all.)

In the book it is clearly given
When base a is any real no and exponent b and c are Integrs under this they stated 6 laws of Exponention

When you stated it, you only said a>0 and a>1 (presumably meaning not equal to 1?), and didn't mention b and c being integers. Some such restriction is needed, if negative bases are to be allowed. The reason is that if the base can be negative, and the exponents can be, say, 1/2, then you start getting complex numbers, and the rules become troublesome.

Now my objective is to find a counter example with base any real no and exponent b and c non integr .
I tried using this eg
(-4)^1/2 =(-4)^1/2 but it works.
(-8)^1/3=(-8)^1/3 but it works
Can u give me one counter example where exponent are non integer with any real base
and the law gets invalid.

Are you sure the first one works? Is (-4)^(1/2) equal to 2i, or -2i?

And although you might think that (-8)^(1/3) is simple enough, once you open the range to complex numbers, as you must if you allow the first example, then it actually has three values!

Why did u said " If b and c can be any real number, then there are problems defining a^b and a^c uniquely" ?

Are u saying that If a=4 and b&c is 1/2 ;

4^(1/2)=4^(1/2)
Then on one side 2 may come and on the right side -2 will come leading to ambiguity.

Yes, that's a big part of it. But it gets far worse, as I said above, if you then allow a to be negative, and more so if you allow the exponents to be irrational.

I will be starting log chapter but what are you saying here . I can use log laws to prove the identity law is invalid or valid ??

When the exponent can be irrational, you need to use logarithms to prove such rules, and what goes wrong is that you'd have to take the logarithm of a negative number, which you can't do.

please give a Eg to understand.

Here's a little English lesson: The abbreviation "e.g." doesn't mean "example" (a noun); rather, it means "for example" (a phrase). You wouldn't say "please give a for example", so you shouldn't use "e.g." as you did here. But we know what you mean.

 

[JeffM]

I tried using this eg
(-4)^1/2 =(-4)^1/2 but it works.
(-8)^1/3=(-8)^1/3 but it works
Can u give me one counter example where exponent are non integer with any real base
and the law gets invalid.

BUT IT DOESN’T WORK

There is no real solution to (-4)^(1/2). If we are dealing with complex numbers

[MATH](- 4)^{(1/2)} = 2i \text { or } - 2i \text { and } 2i \ne - 2i.[/MATH]
So if we are dealing with real numbers, a negative number to an exponent is not always defined at all. To say that one thing that does not exist is equal to a thing that does not exist is meaningless. Moreover, if we are dealing with complex numbers, a negative number to a non-integer exponent is defined, but not uniquely. So the statements of equality may be meaningful, but are false.

[MATH](- 8)^{(1/3)} = - 2, \ 1 + i * \sqrt{3}, \text { or } 1 - i * \sqrt{3}.[/MATH]
And obviously it is false that [MATH]- 2 = 1 + i * \sqrt{3}.[/MATH]
Doubt me?

[MATH](1 + i * \sqrt{3})^3 = (1 + i * \sqrt{3})(1 + i * \sqrt{3})^2 =[/MATH]
[MATH](1 + i * \sqrt{3})\{(1)^2 + 2(1)(i * \sqrt{3}) + (i * \sqrt{3})\} = (1 + i * \sqrt{3})\{1 + 2i * \sqrt{3} + (-1)(3)\} =[/MATH]
[MATH](1 + i \sqrt{3})(- 2 + 2i * \sqrt{3}) = -2 + 2i \sqrt{3} - 2i \sqrt{3} + 2(i * \sqrt{3})^2 = - 2 + 2i^2(\sqrt{3})^2 =[/MATH]
[MATH]- 2 + 2(- 1)(3) = - 2 - 6 = - 8.[/MATH]
Your examples are fallacies.

The fact is that you can work with SOME exponents and a base that is a non-positive real number and make sensible statements, but not all exponents. So the laws of logs do not universally apply to bases that are non-positive real numbers. And you can work sensibly with ANY exponent on a complex base that is not zero, but the results are not unique if the exponents are not integers. So the laws of logs do not apply universally to bases that are complex numbers.

To avoid all these complexities, the laws of exponents are stated for positive real bases, where they universally apply for all real exponents. (I admit, however, that I personally could not prove that they apply when exponents are irrational. In fact, I have never dealt with a practical problem involving irrational exponents.)

Why did u said " If b and c can be any real number, then there are problems defining a^b and a^c uniquely" ?

Are u saying that If a=4 and b&c is 1/2 ;

4^(1/2)=4^(1/2)
Then on one side 2 may come and on the right side -2 will come leading to ambiguity.

But my reasoning is wrong

Yes, you are wrong. It is true that 4 has two square roots, but

[MATH]4^{(1/2)} = 2 = \sqrt{4}[/MATH]
Those functions of non-negative real variables are defined to be non-negative real numbers just to avoid the ambiguity you are talking about.

[MATH]\text {The square roots of 3 are } \sqrt{3} \text { and } - \sqrt{3}.[/MATH]
You are confusing the definitions that give the notation a singular meaning and the multiplicity of even real roots. It is a very easy confusion to fall into.

 

[Saumyojit]

The reason is that if the base can be negative, and the exponents can be, say, 1/2, then you start getting complex numbers, and the rules become troublesome.

I understood that as JeffM also said in post 29 if the base is negative and exponent is fraction then there is no real solution in some cases like -4 but we have solutions only under (under Complex no) then also we don't get unique solutions.

Although cube root of -8 has a unique solution (only considering real no) but for all negative nos in base the law fails to hold.
This is what Dr p meant by trouble some
Right?

That means the laws of Exponention when talking about it ; it is important to consider the domain .
Also i see that (4)^1/2 will not work as its real solutions are not unique although they are defined under real no . Thats another reason why the exponent has to be integers.

It took a lot of playing around with nos to come to a conclusion who genralized that the laws are perfect when the base is real and integer exponent

Yes, that's a big part of it.

Did you said yes to my "wrong reasoning " or
yes to my realization that there will not be unique solutions .
As JeffM says

Yes, you are wrong

I am confused.

Which leads to my another doubt

Those functions of non-negative real variables are defined to be non-negative real numbers just to avoid the ambiguity you are talking about.

The square roots of 3 are √3 and −√3.The square roots of 3 are 3 and −3.\displaystyle \text {The square roots of 3 are } \sqrt{3} \text { and } - \sqrt{3}.

You are confusing the definitions that give the notation a singular meaning and the multiplicity of even real roots. It is a very easy confusion to fall into.

What do you mean by "those functions of non-negative real variables are defined to be non-negative real numbers "

What are those functions?
Even root functions??

What are you telling with this line
"You are confusing the definitions that give the notation a singular meaning and the multiplicity of even real roots."

Are you telling that this definition
√4=2=4^1/2 means it inherently asking for principal root.(Singular meaning)

and this +_√4 means two real roots will come.(Mutliplicity)

something like this??

Also I can't relate to the mathematical notation you(jeffm) showed using the eg of square root 3!

have no idea why the author of your book limited that proposition to integers

u are saying that the author should have expanded the limitations of exponent to irrational Exponents?
I understood that author stated everything keeping in mind all the six laws are valid to the constraint of each variable

 

[Dr.Peterson]

Did you said yes to my "wrong reasoning " or
yes to my realization that there will not be unique solutions .

I was saying yes to what I quoted, which was about the ambiguity.

It appears that JeffM disagrees with me to some extent there; but as I see it, [MATH]4^{1/2}[/MATH] is not identical to [MATH]\sqrt{4}[/MATH], and in particular the rules for exponents applied to the former don't work unless you treat it as double-valued. There are different ways to look at the exact cause of the problem, which is why I said, "that's a big part of it" rather than saying that was exactly the issue.

The radical is defined as the principal root, that is, the non-negative one. The fractional power is (at least sometimes) taken that way too, but that breaks some of the "laws".

 

[JeffM]

I understood that as JeffM also said in post 29 if the base is negative and exponent is fraction then there is no real solution in some cases like -4 but we have solutions only under (under Complex no) then also we don't get unique solutions.

You are correct. The moment that we talk GENERALLY about non-integer exponents and negative bases, we must deal with complex numbers and multiple results of exponentiation. It is true that if the exponent is limited to rational numbers with denominators that are odd integers, we can ignore these complexities because there is a unique real result, but we certainly cannot give any general laws. Your - 8 example is an exception of simplicity in a general world of complexity.

Also i see that (4)^1/2 will not work as its real solutions are not unique although they are defined under real no . Thats another reason why the exponent has to be integers.

It took a lot of playing around with nos to come to a conclusion who genralized that the laws are perfect when the base is real and integer exponent

This is close to correct, but contains subtle errors.

It is true that there are two distinct numbers with a square that is 4, namely 2 and - 2. This creates no problem with respect to treating exponentiation as a function: the function of two different numbers are permitted to have the same result. That fact creates no ambiguity. But a function of a given number must BY DEFINITION give a unique result. Therefore the functions

[MATH]\sqrt{x}[/MATH] and [MATH]x^{1/2}[/MATH] are defined to be non-negative. It is an error to say

[MATH]- 2 = 4^{1/2}[/MATH]. What is true is that [MATH]-2 = - 4^{1/2}[/MATH].

This is what I meant by saying you are confusing the multiplicity of roots, both (-2)^2 and (2)^2 are roots of 4, and the uniqueness implied by the notation 4^(1/2) is 2.

Furthermore, you are wrong when you say the laws are general when the base is real and the exponents are integers. The laws are general when the base is a positive real number and the exponent is a real number. You are allowing the fact that an author limited his discussion to integer exponents to make an unwarranted and false conclusion. I do not know why that author limited his discussion to integer exponents: perhaps to avoid any discussion of irrational numbers.

I think this addresses the question you posed to me.

EDIT: Dr. Peterson posted while I was writing. I see we may disagree on one point. He is probably right and I am probably wrong on that point. He and I will discuss and resolve, but I am out of time right now.

 

[Saumyojit]

I was saying yes to what I quoted, which was about the ambiguity.

But my Eg of using a positive base was wrong .I should have used (-4)^1/2 on both sides to match with what you said "omitting negative base" but also to the power 1/2 if it is double valued I am correct then as 4^1/2 also gives two values.
Then In this particular proposition or law (a^b=a^c) to work my exponent has to be Integer and the base can be any real no . That bottom line.
It's getting more confusing I think which I didn't expected really.

And you say some rules will break .
Can u show some.

, you are wrong when you say the laws are general when the base is real and the exponents are integers. The laws are general when the base is a positive real number and the exponent is a real number

Except this proposition
a^b=a^c =>b=c
I think all the other laws hold valid as JeffM says "base is a positive real number and the exponent is a real number"
I didn't check but I trust him!

 

[JeffM]

Except this proposition
a^b=a^c =>b=c
I think all the other laws hold valid as JeffM says "base is a positive real number and the exponent is a real number"
I didn't check but I trust him!

As I said in an earlier post, my knowledge runs out when it comes to proving things about irrational numbers so there is no reason to trust me on assertions about irrational exponents.

What I have said is my understanding, which is that

All the laws including

[MATH]a^b = a^c \iff b = c[/MATH]
apply when the base is a positive real number and exponents are real. You keep getting caught up in the fact that this one author you read stipulated that the exponents that he is discussing are integers. I do not know why this was done. Perhaps because exponentiation is a closed operation within the integers if both base and exponents are integers. Perhaps because he is interested in discrete mathematics. But the fact that he limited his discussion does not mean that limitation is necessary. Dr. Peterson and I have explained this before.

 

[Dr.Peterson]

But my Eg of using a positive base was wrong .I should have used (-4)^1/2 on both sides to match with what you said "omitting negative base" but also to the power 1/2 if it is double valued I am correct then as 4^1/2 also gives two values.
Then In this particular proposition or law (a^b=a^c) to work my exponent has to be Integer and the base can be any real no . That bottom line.
It's getting more confusing I think which I didn't expected really.

And you say some rules will break .
Can u show some.

Some of what I've said is probably wrong when restricted to real numbers; the trouble is that you have opened the discussion to complex numbers when you allow negative bases, and then you have to take fractional powers as multiple-valued. In general, I want to be very cautious mixing these ideas, which is why I've been trying to avoid absolute statements. I get almost literally dizzy trying to deal with the different contexts you have brought together; I don't want to try to find the maximal conditions under which a given rule applies!

Here is a discussion I like about some of the rules that break (including some talk about multiple values): Does (a^m)^n Always Equal a^(mn)?

 

[Saumyojit]

When n is a rational number
with (least positive) denominator d, then x^n has d different complex Values

what is the least positive Denominator 'd' ?

for me if n is a fractional no then the numerator and denominator has to be in lowest terms .There can be 3 combinations
(odd numerator and denominator even;
odd numerator and denominator both ;
even numerator and denominator odd).

is he saying like if i have 3 ^ 2/3 then the cube root of the radicand will yield 3 different non real or real roots

 

[Dr.Peterson]

what is the least positive Denominator 'd' ?

for me if n is a fractional no then the numerator and denominator has to be in lowest terms .There can be 3 combinations
(odd numerator and denominator even;
odd numerator and denominator both ;
even numerator and denominator odd).

is he saying like if i have 3 ^ 2/3 then the cube root of the radicand will yield 3 different non real or real roots

Please state what you are quoting! This is apparently from the page I referred to, which said this:

The "rule"

(a^b)^c = a^(b*c),

which works so nicely for integer exponents and even for non-integer
exponents when the base is positive, becomes problematic with negative
or complex bases and non-integer exponents. You see, when n is an
integer, then x^n has only one value, but when n is a rational number
with (least positive) denominator d, then x^n has d different complex
values, and when n is irrational, then x^n has infinitely many
different complex values. The rule (a^b)^c = a^(b*c) should be
interpreted as: Every value of a^(b*c) is one of the values of
(a^b)^c. (It doesn't even work in the other direction, since
sometimes some of the values of (a^b)^c aren't any of the values of
a^(b*c). For example, -1 is one of the square roots of 1^2.)

First, no, fraction does not always have to be written in lowest terms; but here he is assuming it is, which is indeed what he means by least positive denominator.

Second, yes, he is saying that 3^(2/3), for example, has three values. Here they are:

 

[Saumyojit]

That means if the fraction was given like this 4/6 then 6 cannot be the least Denom bcoz they have Common factor so 2/3 is right one where 3 is least positive Denominator.
Ok ?

 

[Dr.Peterson]

That means if the fraction was given like this 4/6 then 6 cannot be the least Denom bcoz they have Common factor so 2/3 is right one where 3 is least positive Denominator.
Ok ?

Yes.

 

[Saumyojit]

Some of what I've said is probably wrong when restricted to real numbers

Which statements are you referring to ?
I think you had told every thing carefully i.e "when we will use negative base we deal with complex no " in post 28
Which other parts are u referring to?

What is the fuss about the ^1/2 notation.
I saw that when the no or base is negative the law broke and also by talking the principal root of the radicand .I think we discussed about this law and its invalidity in this post if you go back.
But I don't get the point what is the connection if I don't take 1/2 as double valued with the post in link

Is the root single valued or mutlivalued function.
I saw in wiki we may consider sqrt√ as multivalued .
Context is important right??
Finding roots of a polynomial, then I should have both +_ but only√4 then principal square root we talk about.