- #1

Bashyboy

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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?

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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