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Spivak Chapter 1 Problem 1 (i) is my proof correct?

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1. Spivak 4th Edition Problem 1 (i) Chapter 1
If ax = a for some a not equal to 0, then x = 1.

2. P7 where a*a^-1=a^-1*a=1

3. Using P7
Then (x^-1)ax=a
then (X^-1 * x)a=a
then 1*a=a
then x=1

Am I approaching this correct or am I supposed to prove P7 as well or prove this a whole different way. Thank you in advance.
 
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  • #2
Fredrik
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2. P7 where a*a^-1=a^-1*a=1
To be more precise: For all non-zero real numbers a, there's a real number a-1 such that a-1a=aa-1=1. (We are talking about real numbers, right?)

Then (x^-1)ax=a
This doesn't follow from P7 or from the details given in the problem statement, unless you have assumed that x=1, and that's what you're supposed to prove.

1*a=a
then x=1
Here you're saying that since 1a=a implies x=1. This would mean that all real numbers are =1 (or that 1a≠a).

What you need to do is to use P7, in the form I stated it, together with the assumptions in the problem statement.

am I supposed to prove P7 as well
If P7 is what i said above (after "to be more precise"), then no. P7 is a part of the definition of "real number". There are mathematical objects that don't have that property, but we wouldn't call them real numbers.
 
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To be more precise: For all non-zero real numbers a, there's a real number a-1 such that a-1a=aa-1=1. (We are talking about real numbers, right?)


This doesn't follow from P7 or from the details given in the problem statement, unless you have assumed that x=1, and that's what you're supposed to prove.


Here you're saying that since 1a=a implies x=1. This would mean that all real numbers are =1 (or that 1a≠a).

What you need to do is to use P7, in the form I stated it, together with the assumptions in the problem statement.
Thanks! Yes, it was for real numbers. I was assuming x=1 and using x in place of a when I used P7.

Again the problem was "If ax = a for some a not equal to 0, then x = 1."

Assume x = 1 and
Then by definition of P7 (x*x^-1)a=a
then 1*a=a
Then x=1

Is that still incorrect? I apologize I am just starting proofs.....thank you for your help.
 
  • #4
HallsofIvy
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1. Spivak 4th Edition Problem 1 (i) Chapter 1
If ax = a for some a not equal to 0, then x = 1.
First, what is the setting here? Are these numbers, or members of an abstract field?

2. P7 where a*a^-1=a^-1*a=1
Okay, so, in either case, you have the fact that there exist multiplicative inverses. This would be better expressed, "If a is not 0 then there exist a^-1 such that a*a^-1= a^-1*a= 1". Isn't that what Spivak actually gives as "P7"?

3. Using P7
Then (x^-1)ax=a
No, you can't do this! You dont know, yet, that x is not 0 and so do not know if it has an inverse! Since you are specifically told that a is not 0, and so has an inverse, how about using a^-1 here, rather than x^-1?

then (X^-1 * x)a=a
then 1*a=a
then x=1

Am I approaching this correct or am I supposed to prove P7 as well or prove this a whole different way. Thank you in advance.
I don't know whether, as I said before, if this is "the real numbers" (or "the rational numbers") specifically or an abstract field. If the former, you probably could prove it from whatever definition you are using for the real numbers (or rational numbers) but that, I suspect, would be very difficult. If the latter, I expect that "P7" is given as a postulate and you do not prove postulates.
 
  • #5
HallsofIvy
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Thanks! Yes, it was for real numbers. I was assuming x=1 and using x in place of a when I used P7.

Again the problem was "If ax = a for some a not equal to 0, then x = 1."

Assume x = 1 and
No, you want to prove x= 1. You cannot start by assuming what you want to prove! You are told that a is not 0 so you know a has an inverse. Use a^-1 rather than x^-1 in your first proof.

Then by definition of P7 (x*x^-1)a=a
then 1*a=a
Then x=1
So you arrive at what you assumed is that surprise?

Is that still incorrect? I apologize I am just starting proofs.....thank you for your help.
Don't apologize for being wrong. Keep trying and eventually you will be right.
 
  • #6
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No, you want to prove x= 1. You cannot start by assuming what you want to prove! You are told that a is not 0 so you know a has an inverse. Use a^-1 rather than x^-1 in your first proof.


So you arrive at what you assumed is that surprise?


Don't apologize for being wrong. Keep trying and eventually you will be right.
OK So is this correct then:

We are given a is not equal to zero and is a real number.
We are given P7 (Existence of multiplicative inverses) a*a^-1=a^-1*a=1, for a not equal to 0
Then (a*a^-1)x=a*a^-1
Then 1*x=1
Then x=1

Thank you!
 
  • #7
HallsofIvy
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Yes, that is exactly right!:biggrin:
 
  • #8
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Thanks so much for the help!
 
  • #9
Fredrik
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OK So is this correct then:

We are given a is not equal to zero and is a real number.
We are given P7 (Existence of multiplicative inverses) a*a^-1=a^-1*a=1, for a not equal to 0
Then (a*a^-1)x=a*a^-1
Then 1*x=1
Then x=1

Thank you!
As HallsofIvy said, you're doing it right this time. But you left out something important in the first step: You didn't actually say that you're using the equality you had been given: ax=a.

Since you have already posted a correct solution, I don't mind posting my solution:

(1) ax=a (This is what you were given).
(2) a-1(ax)=a-1a (Since a≠0, a has a multiplicative inverse. We got this equality by multiplying both sides of the previous one by a-1 from the left).
(3) (a-1a)x=a-1a (Multiplication is associative). Edit: I wrote "commutative" here first...brain fart.
(4) 1x=1 (Property of multiplicative inverses).
(5) x=1 (Property of 1).
 
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Thanks for your help Fredrik, very helpful....would it be proper to always number my steps? As well your bracketed comments, should I be including something each time?
 
  • #11
HallsofIvy
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Hey, you don't want to give your teacher a heart attack!:wink:

I wouldn't say you should number your steps- but writing exactly why you doing each step would be excellent. (And amaze your teacher as I suggested!)
 
  • #12
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I would prefer:

ax=a
ax-a=0
a(x-1)=0
which implies that a=0 or x-1=0 (since R contains no zero divisors). Since a is given as non-zero we conclude that x-1=0 which gives x=1.

I should note that this proof avoids the use inverses and thus proves this result not only for any field but for any integral domain as well.
 
  • #13
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Geez....can not believe I did not see that way of proving it. Thanks so much very helpful!

Is the best way to learn proving just plugging along through a Spivak or Apostol or should I be doing something else first?

Thank you all for your help.
 

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