View Feed
Coffee Room
Discuss anything here - everything that you wish to discuss with fellow engineers.
12933 Members
Join this group to post and comment.
tashirosgt • Jul 19, 2009

(-1)^(2/3) = 1 ?

Most web pages about secondary school algebra that I have seen are logically inconsistent when it comes to exponentiation. On the one hand, they claim (x [sup] a [/sup]) [sup] b [/sup] = x [sup] (ab) [/sup] whenever the exponentiations are defined. Then they allow operations like (-1)[sup] (2/3) [/sup] to be defined and usually give examples of how to work such things out (integer powers of odd roots of negative numbers).. So if you let x =- 1, a = 2/3 and b = 3/2 in the above law, you get the contradiction that 1 = -1.

Before anyone decides to wax eloquent on the properties of the complex numbers, the cube roots of unity etc. , let me say that I'm talking about whether these web pages use a consistent set of assumptions for the real numbers. The complex numbers are interesting, but they are no excuse for presenting an inconsistent set of axioms for the real numbers.
And I'm talking about exponentiation as an operation that is supposed to produce at most one result, not as a notation for the set of several roots to an equation.

It's particularly interesting that many web pages state that x [sup] a [/sup] is only defined when x is a positive number. But later, they cannot resist showing students examples of how to compute the odd roots of negative numbers. As to whether x [sup] (6/10) [/sup] = x [sup] (3/5) [/sup] when x < 0, most pages avoid that issue.

Mathematical software and calculators are also inconsistent. Try computing (-2)[sup] (0.6) [/sup] with various calculators. There is some excuse for that. They are trying to read the users mind to determine if he wants to stay in the real number system.

I think the correct treatment of the above issues has to do with correctly distinguishing between the operation of exponentiation (which obeys "the laws of exponents") and a common extension of that operation, which doesn't.
shalini_goel14 • Jul 19, 2009
Hey tashirosgt,

Good observation kept before. Can you tell the reason clearly behind it, why it happens so? Is there any rule/logic made by our old mathematicians for this ? 😕
Kaustubh Katdare
Kaustubh Katdare • Jul 19, 2009
Ha! Looks interesting. I'll look into this later today.

[PS: Thread moved to mathematics section]
tashirosgt • Jul 20, 2009

I haven't studied the history of mathematics very much so I don't how the truly old mathematicians thought about this subject (for example the ancient Greeks). The thinking of modern mathematicians (e.g. "Principles Of Mathematical Analysis" Second Edition (1964) , by Walter Rubin) is that the simplest way to do things (in the axioms for the real numbers) is to give a definition for x [sup](1/ n} [/sup] when n is a positive integer and x > 0. Nothing is defined about the case x < 0, so exponentiation by 1/n remains undefined in that situation. Then there is no problem with the law (x [sup] a [/sup] ) [sup] b [/sup] = x [sup] (ab) [/sup] since this law is asserted only in the case when all the exponential operations are defined.

Rudin's book is a college text. We see that at a secondary school level, the authors of books and web pages are not so careful about logical consistency.

People's confusion about (-1)[sup] (2/3} [/sup] can come from a very old fashioned idea about mathematics. This is the idea that a string of symbols has some "true" meaning and that a mathematical definition is merely an explanation of that "true" meaning expressed in words. The modern view would be that definitions are arbitrary and established by custom. They are not "true" or "false". The interpretation of the a string of symbols comes from the mathematical definition, not vice versa. It is permissible to give a mathematical definition for a string of symbols that extends an existing definition, but you can't assume the laws that applied to the symbols in the old definition will automatically apply to the symbols in the new definition.
Good Discussion 😀

My question very general but can some one solve my doubt.

Why - * - = + ??
minus*minus = Plus , but why?
Good Discussion 😀

My question very general but can some one solve my doubt.

Why - * - = + ??
minus*minus = Plus , but why?
There is a number line concept explaining this !!!

IS anyother explanation also there??
tashirosgt • Jul 20, 2009
To prove (-A)(-B) = AB rigorously, you must state the axioms for the real numbers and prove various other things. But the rough idea is:

Suppose the following facts are true for all real numbers A,B,C:
A(B + C) = AB + AC
A(0) = 0
B + (-B) = B - B = 0
A(-B) = -(AB)

The you must have (-A)(0) = 0 = (-A) (B + (- B)) = (-A)B + (-A)(-B) = -AB + (-A)(-B) = 0

The quantity that when added to -(AB) produces 0 is the additive inverse of AB
So (-A)(-B) is the additive inverse of (-AB). The addtive inverse of -AB is AB. So (-A)(-B) must be AB.

(This assumes you have previously established that a number has one and only one additive inverse.)

Strictly speaking, you must distinguish betwen the phrase "a minus" and the phrase "a negative number". If X = -7 then -X is a positive number.
Thanks for the explanation dude 😀

There is one more number line explanation too..

If we want 2 * -3

The expression -3 says starting at zero, go left 3 steps. The expression 2 x means do it twice.

Where do you finish? -6

If you see why you finish on the left side, you have your answer.

What about minus x minus. Let's calculate (-2) x (-3).

Again -3 says starting at zero go left 3 steps. The expression -2 x means, no, go the other way instead -- twice.

Where do you finish? +6. Minus x minus = plus.

Source:Wiki Answers.

Share this content on your social channels -