Spivak Example

  • Thread starter Bashyboy
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  • #1
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Hello,

I am embarking to read Spivak's book on Calculus, and have come across some difficulty with something that is perhaps rather trivial. In the third edition, there is a section entitled Basic Properties of Numbers. Near the end of page 7, the author begins discussing how he will use property 9 to derive the fact that, when negative numbers are multiplied, the result is a positive number. Before this, he must show that (-a) x b = -(a x b) is true. To prove this, we must show that
(-a) x b + a x b = [(-a) + a] x b, which I can follow. Next, he says that, because this is true, then we can add -(a x b) to both sides of the equation:

[itex](-a) \cdot b + a \cdot b + [-(a \cdot b)] = [(-a) + a] \cdot b + [-(a \cdot b)][/itex]

Clearly, the first term on the RHS of the equation will yield zero.

[itex](-a) \cdot b + a \cdot b + [-(a \cdot b)] = -(a \cdot b)[/itex]

[itex][-a + a - a] \cdot b = -(a \cdot b)[/itex]

[itex](-a) \cdot = -(a \cdot b)[/itex]

But are these steps truly valid; how I can know that, when factoring the b, the negative symbol isn't appended to the b; isn't this what I am trying to prove, that the negative symbol is appended to the a?
 
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Answers and Replies

  • #2
575
76
Hello,

I am embarking to read Spivak's book on Calculus, and have come across some difficulty with something that is perhaps rather trivial. In the third edition, there is a section entitled Basic Properties of Numbers. Near the end of page 7, the author begins discussing how he will use property 9 to derive the fact that, when negative numbers are multiplied, the result is a positive number. Before this, he must show that (-a) x b = -(a x b) is true. To prove this, we must show that
(-a) x b + a x b = [(-a) + a] x b, which I can follow. Next, he says that, because this is true, then we can add -(a x b) to both sides of the equation:

[itex](-a) \cdot b + a \cdot b + [-(a \cdot b)] = [(-a) + a] \cdot b + [-(a \cdot b)][/itex]

Clearly, the first term on the RHS of the equation will yield zero.

[itex](-a) \cdot b + a \cdot b + [-(a \cdot b)] = -(a \cdot b)[/itex]

[itex][-a + a + a] \cdot b = -(a \cdot b)[/itex]

[itex](-a) \cdot = -(a \cdot b)[/itex]

But are these steps truly valid; how I can know that, when factoring the b, the negative symbol isn't appended to the b; isn't this what I am trying to prove, that the negative symbol is appended to the a?

Are those the actual steps listed in the text? The next-to-last line looks a bit suspect. It seems like it should be

##(-a)\cdot b+a\cdot b=(-a+a)\cdot b## by distributivity of multiplication over addition
##(-a)\cdot b+a\cdot b=0\cdot b## by definition of additive inverses
##(-a)\cdot b+a\cdot b=0## from results related to the definition of the additive identity and distributivity of multiplication over addition
##[(-a)\cdot b+a\cdot b]+(-(a\cdot b))=0+(-(a\cdot b))## because addition is well-defined as a binary operation (?)
##[(-a)\cdot b+a\cdot b]+(-(a\cdot b))=-(a\cdot b)## by definition of additive identity
##(-a)\cdot b+[a\cdot b+(-(a\cdot b))]=-(a\cdot b)## by associativity of addition
##(-a)\cdot b+0=-(a\cdot b)## by definition of additive inverse
##(-a)\cdot b=-(a\cdot b)## by definition of additive identity

It's possible that some of those steps were left out, but that's what is going on as far as I can tell.
 
  • #3
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Oh, yes, I see now. Thank you very much.
 

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