Homework Check: Proving Equations using Exponent Rules

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Yizewu

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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!
 
I think in 2b a must not be zero or you would be dividing by zero which is undefined. On the other hand for 2d if m was 2 and a was -1 you will get the square root of -1
 
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
 
Yizewu said:
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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!
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Technically, none of the facts in problem 2 can be proved, because you only have a definition for positive integer powers, not for zero, negative, or fractional exponents! What they are really asking you to do is to show that these are definitions that are consistent with the assumption that the properties in problem 1 extend to such exponents.

You have the right ideas for each question, but the way you showed your work doesn't really show the desired conclusion. For example, for (b), you never actually wrote "1/a^n" in your work!

I don't see that you ever "add exponent m to one side of the equation without adding it to the other". Where do you think you did that?

AbdelRahmanShady said:
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
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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.
 
@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.
 
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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
 
AbdelRahmanShady said:
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
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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.

And 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.
 
JeffM said:
@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.
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My main objection was the use of symbol \(\displaystyle \sqrt{ } \) without explanation (none in problem-1)
 
Subhotosh Khan said:
My main objection was the use of symbol \(\displaystyle \sqrt{}\) without explanation (none in problem-1)
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Yes, you need a definition of both the radical symbol and an exponent that is the multiplicative inverse of a non-zero integer.
 
In 2a, they imply that the base can be zero. 0^0 is undefined, not 1.

In 4, they imply that the base in 2c cannot be zero. The mth root of zero is 0.

Send that exercise to its room without supper.

:rolleyes:
 
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.
Dr.Peterson said:
And 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.
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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
 
AbdelRahmanShady said:
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
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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.

You're certainly right that if you accept the existence of a^0, then it's proved. The OP did so, just without quite saying everything right. We haven't denied that. We're talking about deeper matters, which the author of the problem should not have glossed over if the whole point is to introduce the idea of rigor. Once you accept the idea that we are really defining powers as whatever you can obtain from the original definition, extending it to cases that as yet have no definition, then it's all good (though you have to be careful with other details, some of which have been mentioned).

I could feel that I'm being too picky, except that a similar issue arises in calculus. You can manipulate a limit to obtain all sorts of results, but without ever noticing that the limit doesn't exist. Until you prove it exists, none of the work you do is valid. Here, it is conceivable that it could turn out that a^0, or a^-1, or a^(1/2), is not properly defined because of some inconsistency. (In fact, we come very close with a^(1/2), which needs extra conditions in order to be uniquely defined, which hasn't been done here.)

By the way, you need to learn to use parentheses carefully. You didn't really mean a^m-n, but rather a^(m-n). As you wrote it, n is not in the exponent.
 
AbdelRahmanShady said:
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
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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 then

[math]n = m \text { and } a \ne 0 \implies \dfrac{a^n}{a^m} = 1.[/math]
But you cannot prove that therefore [imath]a^0 = 1[/imath] because the definition in 1 is clearly meaningless with respect to any non-positive number.

In fact the definition in 1 is insufficient to show that [imath]a^1 = a[/imath] without a definition of why a fits the product definition given in 1.

Personally, I like to define non-negative integer exponents as

[math]n \text { a non-negative integer and } r > 0 \implies \\ r^n = 1 \text { if } n = 0 \text { and}\\ r^n = r * r^{(n-1)} \text { if } n > 0.[/math]
That does not reflect the historical origins of the notation, but it simplifies the extension of the notation to exponents that are negative integers or rational numbers.
 
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
 
AbdelRahmanShady said:
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
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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 is

[math]m = n \implies \dfrac{a^n}{a^m} = 1.[/math]
It is a very simple concept: zero is not a positive integer and thus [imath]a^0[/imath] is not defined. We need a different definition.
 
AbdelRahmanShady said:
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
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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.)

But anyone at a higher level of mathematics and doing formal proofs needs to pay attention to such details. To see an extended presentation of the idea that the basic definition is not merely applied to new kinds of numbers, but extended in a compatible way, see Wikipedia. (See under Zero exponent, Negative exponents, Rational exponents, ...)
 
Because it defines [imath]a^m[/imath] subject to the constraint that m must be a positive integer, and zero is not a positive integer. Positive integers are defined as integers greater than zero.
 
Yes but definition doesn’t say a^x x can’t be zero it says that m and n are element of z positive and m can be equal n
 
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