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ActionPotential
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I am taking Calculus at the end of the month for the first time but I have started working through Spivak (wanted to get a headstart and didn't have the class textbook yet) and I have been grasping his beautiful description of some the properties of real numbers yet there is a very minor detail in P9 in Chapter 1 in which a step is missing (or as I realized as I thought about it, I was missing something). I just need clarification to make sure I am understanding what happened.
(P9) If a, b, and c are any numbers, then[itex]a[/itex][itex]\cdot[/itex][itex](b+c)=a[/itex][itex]\cdot[/itex][itex]b+a\cdot[/itex]c
If [itex]a-b=b-a[/itex]
then [itex](a-b)+b=(b-a)+b=b+(b-a)[/itex]
hence [itex]a=b+b-a[/itex]
hence [itex]a+a=(b+b-a)+a=b+b[/itex]
Consequently [itex]a[/itex][itex]\cdot[/itex][itex](1+1)= b[/itex][itex]\cdot[/itex][itex](1+1)[/itex]
and therefore [itex]a=b[/itex]
Now does a+a=b+b become a(1+1)=b(1+1) because of p9 [itex]a[/itex][itex]\cdot[/itex][itex](b+c)=a[/itex][itex]\cdot[/itex][itex]b+a\cdot[/itex]c?
I know this should be incredibly obvious and once I understand it I will kick myself but I couldn't move on without knowing exactly.
(P9) If a, b, and c are any numbers, then[itex]a[/itex][itex]\cdot[/itex][itex](b+c)=a[/itex][itex]\cdot[/itex][itex]b+a\cdot[/itex]c
If [itex]a-b=b-a[/itex]
then [itex](a-b)+b=(b-a)+b=b+(b-a)[/itex]
hence [itex]a=b+b-a[/itex]
hence [itex]a+a=(b+b-a)+a=b+b[/itex]
Consequently [itex]a[/itex][itex]\cdot[/itex][itex](1+1)= b[/itex][itex]\cdot[/itex][itex](1+1)[/itex]
and therefore [itex]a=b[/itex]
Now does a+a=b+b become a(1+1)=b(1+1) because of p9 [itex]a[/itex][itex]\cdot[/itex][itex](b+c)=a[/itex][itex]\cdot[/itex][itex]b+a\cdot[/itex]c?
I know this should be incredibly obvious and once I understand it I will kick myself but I couldn't move on without knowing exactly.
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