Spivak's Calculus Prologue Problem 3 (v)

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In summary, the conversation discusses a proof for the equation \frac{a}{b}/\frac{c}{d} = \frac{ad}{bc}, and the importance of presenting the proof in a clear and organized manner. The conversation also touches on the concept of inverses and the different approaches to solving a proof.
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Physics2341313
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Here's my attempt at this proof. Is this correct?


Homework Statement



Prove [itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex] = [itex]\frac{ad}{bc}[/itex]

Homework Equations



P 1-12

The Attempt at a Solution



[itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex] = [itex]\frac{ad}{bc}[/itex]

[itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex] = (ad)(bc)[itex]^{-1}[/itex]

[itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex] = (ad)(b[itex]^{-1}[/itex]c[itex]^{-1}[/itex])

[itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex] = (ab[itex]^{-1}[/itex])(dc[itex]^{-1}[/itex])

[itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex] = (ab[itex]^{-1}[/itex])(d[itex]^{-1}[/itex]c)[itex]^{-1}[/itex]

[itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex] = [itex]\frac{a}{b}[/itex][itex]/[/itex][itex]\frac{c}{d}[/itex]



Also, do proofs have to be in if, then, hence form like they are when Spivak is presenting the basic properties of numbers?
 
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  • #2
The proof can be in any format as long as your steps are clear.

Your steps in this proof are clear, and you are correct. However, there are two things for you to consider.

1. You started with the answer and worked backwards to a statement that had to be true. This is okay as long as all your implications (steps) go in both directions, so that you could have started at the bottom and worked your way up to the top correctly. In fact, the best presentation of this proof is to do that. Start with a known fact or facts and work your way to the conclusion you want.

2. You took a long way around to do this. Do you remember in 5th grade arithmetic when they taught you things like 3/4 divided by 4/5 is 3/4 x 5/4? Is that not true for every number? So can't you just say ##\frac{a/b}{c/d}## = a/b x d/c = ad/bc (as per your studies in the 5th grade). Or did they explicitly want you to use negative exponents?
 
  • #3
Ok thank you that cleared up a few questions I had concerning proofs.

Yes, I had thought about doing that but I didn't know if that would be considered "rigorous" enough for a proof. Spivak's text is my first encounter with any rigorous math and I assumed my proof could only be based on the properties presented earlier in the chapter/prologue - Spivak presents division as being defined in terms of multiplication of a negative exponent after listing the property for a multiplicative inverse. That's why I stuck directly to the negative exponents
 
  • #4
Physics2341313 said:
Ok thank you that cleared up a few questions I had concerning proofs.

Yes, I had thought about doing that but I didn't know if that would be considered "rigorous" enough for a proof. Spivak's text is my first encounter with any rigorous math and I assumed my proof could only be based on the properties presented earlier in the chapter/prologue - Spivak presents division as being defined in terms of multiplication of a negative exponent after listing the property for a multiplicative inverse. That's why I stuck directly to the negative exponents

Yes, for a course you should always do things the way they tell you.

I imagine Spivak said said ##x\cdot x^{-1}## (x inverse) has to be 1, so ##x^{-1}## has to be 1/x. He is probably trying to start in on the general idea of inverses.

You can always ask if you think your book or professor has done something wrong or created a harder way than necessary. It's a good way to learn more.

That all said, it's not how I would have done it. But then, I never wrote that kind of book.
 

1. What is "Spivak's Calculus Prologue Problem 3 (v)"?

"Spivak's Calculus Prologue Problem 3 (v)" is a problem from the book "Calculus" written by Michael Spivak. It is a part of the prologue section and is the third problem in that section.

2. What is the difficulty level of this problem?

This problem is considered to be of advanced difficulty level and is typically tackled by students who have a strong foundation in calculus and mathematical proofs.

3. Can you explain the problem statement?

The problem statement asks to prove that for any real numbers x and y, if x is less than y, then there exists a rational number q such that x is less than q and q is less than y.

4. What is the significance of this problem?

This problem is significant because it introduces the concept of rational numbers and their relationship to real numbers. It also requires students to use their understanding of inequalities and proofs.

5. How can one approach solving this problem?

One approach to solving this problem is to start by understanding the definitions of real and rational numbers. Then, use the fact that x is less than y to construct a rational number q that falls between them. This can be done by using the Archimedian property of real numbers.

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