So, I've been going through Spivak's Calculus, section I, and I'm having some trouble proving 1.1.iii. Also, I'd like to check 1.1.ii and ensure that I did 1.1.i right. So, first, I'm restricted to these nine properties:(adsbygoogle = window.adsbygoogle || []).push({});

a+(b+c)=(a+b)+c <P1, addition associative>

a+0=0+a=a <P2, additive identity>

a+(-a)=(-a)+a=0 <P3, additive inverse>

a+b=b+a <P4, addition commutative>

a*(b*c)=(a*b)*c <P5, multiplication associative>

a*1=1*a=a <P6, multiplicative identity>

a*a^{-1}=a^{-1}*a=1 <P7, multiplicative inverse>

a*b=b*a <P8, multiplication commutative>

a*(b+c)=a*b+a*c <P9, multiplication distributive>

P7 has the condition that a cannot be 0.

So, the first question asks me to prove that if a*x=a, then x=1. So, I start with a*x=a.

a*x=a

a^{-1}*(a*x)=a^{-1}*a <Multiply by a^{-1}>

(a^{-1}*a)*x=a^{-1}*a <P5>

1*x=1 <P7>

x=1 <P6>

Now that one I'm confident on. The second one I know I got right, but I may have done It the wrong way. The second problem asks me to prove that x^{2}-y^{2}=(x-y)(x+y). So...

x^{2}-y^{2}=(x-y)(x+y)

=(x-y)*x+(x-y)*y <P9>

=(x*x-x*y)+(x*y-y*y) <P9>

=((x^{2}+(-x*y))+(x*y))-y^{2}<P1>

=(x^{2}+((-x*y)+(x*y)))-y^{2}<P1>

=(x^{2}+0)-y^{2}<P3>

=x^{2}-y^{2}<P2>

So, I got it right, but was I supposed to modify the left side instead? Does it matter?

Now, the third one has me lost. It asks me to prove that if x^{2}=y^{2}, then x=y or x=-y. I tried. Here's how far I got.

x*x=y*y

x^{-1}*(x*x)=x^{-1}*(y*y) <Multiply by x^{-1}>

(x^{-1}*x)*x=(x^{-1}*y)*y <P5>

1*x=(x^{-1}*y)*y <P7>

x=(x^{-1}*y)*y <P6>

y^{-1}*x=((x^{-1}*y)*y)*y^{-1}<Multiply by y^{-1}>

y^{-1}*x=(x^{-1}*y)*(y*y^{-1}) <P5>

y^{-1}*x=(x^{-1}*y)*1 <P7>

y^{-1}*x=x^{-1}*y <P6>

I don't know where to go from there without recycling the original equation. If only I could just raise both sides of the original equation to the power of 1/2... but I can't...

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# Homework Help: Spivak's Calculus Section I Trouble

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