- #1
Char. Limit
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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:
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 x2-y2=(x-y)(x+y). So...
x2-y2=(x-y)(x+y)
=(x-y)*x+(x-y)*y <P9>
=(x*x-x*y)+(x*y-y*y) <P9>
=((x2+(-x*y))+(x*y))-y2 <P1>
=(x2+((-x*y)+(x*y)))-y2 <P1>
=(x2+0)-y2 <P3>
=x2-y2 <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 x2=y2, 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...
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 x2-y2=(x-y)(x+y). So...
x2-y2=(x-y)(x+y)
=(x-y)*x+(x-y)*y <P9>
=(x*x-x*y)+(x*y-y*y) <P9>
=((x2+(-x*y))+(x*y))-y2 <P1>
=(x2+((-x*y)+(x*y)))-y2 <P1>
=(x2+0)-y2 <P3>
=x2-y2 <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 x2=y2, 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...