I don't think 2(c) can be proven from 1.View attachment 28656
View attachment 28657
Are these questions correct? I am quite confused about 2c and 2d as I wasn't sure if you could add exponent m to one side of the equation without adding it to the other as that would completely change the equation. If it's wrong, can anyone help with the problem?
Thank you!
View attachment 28656
View attachment 28657
Are these questions correct? I am quite confused about 2c and 2d as I wasn't sure if you could add exponent m to one side of the equation without adding it to the other as that would completely change the equation. If it's wrong, can anyone help with the problem?
Thank you!
The main issue will be that any number has two square roots, and for negative numbers this results in inconsistency. Just try taking the square root of the square of a negative number, and compare with the claimed property.Can you clarify what inconsistencies mean do you mean logic problem as contradiction or something like square root of -1 as you are working with reals not complex numbers
But 1b is not presented as a definition; it's proved as a property that applies only to positive integers, because that is true of the definition on which it is based.Can’t you just use definition 1 b to prove 2a like a^m / a^m by definition 2a it is equal a ^m-m which is a ^ 0 so a^0 equal a^m/a^m which is equal 1 if a^m is not zero
My main objection was the use of symbol \(\displaystyle \sqrt{ } \) without explanation (none in problem-1)@Subhotosh Khan
You are correct that we cannot rigorously prove 2c from the definition given in 1, but we cannot rigorously prove 2a or 2b either. The reason is that m and n are limited in 1 to positive integers. There is as yet no definition for an exponent that is a non-negative. I think what the question means by “show” is to demonstrate that expanding the definition of exponent by using 2a, 2b, and 2c as supplementary definitions results in a logically consistent and useful system that covers all rational numbers.
Sloppy language confuses the better students.
EDIT: I should have acknowledged Dr. Peterson’s response in the main body of my response because he was the first to point out the deficiency of the definition in 1 for purposes of 2. I apologize, but my excuse is that I was focusing on what I thought was a useful exercise marred by sloppy language.
Yes, you need a definition of both the radical symbol and an exponent that is the multiplicative inverse of a non-zero integer.My main objection was the use of symbol \(\displaystyle \sqrt{}\) without explanation (none in problem-1)
No I think 2a can be proved rigorouslyAnd my claim is that 2a, rather than being a mere property as stated, is in fact a definition, so it can't be proved; it can only be proved to be consistent with the properties previously shown to be true in the restricted case.
I think you're missing my point. You can't even talk about a^0 if the only definition of powers that exists applies only to positive integers.But 1b is not presented as a definition; it's proved as a property that applies only to positive integers, because that is true of the definition on which it is based.
No I think 2a can be proved rigorously
We know a^m / a^n = a^m-n if m>= n
So I can set n = m. Then a^m / a^m = a^0 moreover a^m / a^m = 1 so a^0 = 1. I just used property 1a which had to be proven at that point
NO. IT IS NOT. The definition in 1 carefully restricts exponents to POSITIVE integers. You can prove that if n and m are both positive integers thenBut 1b is not presented as a definition; it's proved as a property that applies only to positive integers, because that is true of the definition on which it is based.
No I think 2a can be proved rigorously
We know a^m / a^n = a^m-n if m>= n
So I can set n = m. Then a^m / a^m = a^0 moreover a^m / a^m = 1 so a^0 = 1. I just used property 1a which had to be proven at that point
This is the last I am going to say about your error. The definition in 1 does not permit an exponent of 0. You are correct that the definition permits m = n. Therefore all you can say USING THE DEFINITION IN 1 isNo because m can be equal to n so we know a^0 = a^m/a^m and we know that x/x = 1 if x is not zero so if a is not zero a^0 = 1
I'll just repeat what I've said: Your algebra is correct, and is undoubtedly what the author of the problem expects! We are just saying that the author is overlooking a basic issue of logic. Apparently at the level of his text, he feels it is not necessary to be that precise; and we don't know anything more about the book. (Possibly he even says some of these things elsewhere in the context.)No because m can be equal to n so we know a^0 = a^m/a^m and we know that x/x = 1 if x is not zero so if a is not zero a^0 = 1