Prove that if ab=ac, then b=c.

[kripkrip420]

Can someone help me prove that if ab=ac and a does not equal 0, then b=c. You can only use the field axioms. Here is my attempt( I am not sure about the final result).

ab=ac
then,
ab-ac=0
a(b-c)=0
a(0)=0 (Already proved this)
since a cannot equal 0, (b-c) must equal zero and we are left with...
b-c=0
b=c

The only problem I have found with the above is that I have assumed that a does not equal 0. I do not know if this is alright since the statement "a does not equal 0" is part of the originial theorem I am trying to prove.

Thank you to anyone who helps!

 

[chiro]

There is nothing wrong with your proof. The assumption that a <> 0 needs to hold to show b = c. If a was zero then a solution b <> c would exist which is why they stated that a must be non-zero.

 

[gb7nash]

And more importantly, every field is an integral domain, so b-c must equal 0.

 

[I like Serena]

if ab=ac and a does not equal 0, then b=c. You can only use the field axioms.

The only problem I have found with the above is that I have assumed that a does not equal 0. I do not know if this is alright since the statement "a does not equal 0" is part of the originial theorem I am trying to prove.

Thank you to anyone who helps!

The proof starts with the two initial statements:
1. ab=ac
2. a does not equal 0

The proof requires that using these 2 statements, combined with the field axioms, you deduce that:
3. b=c

You did exactly that, and so yes, you can use the statement "a does not equal 0".
What you cannot do, is use the statement that b=c, and you didn't.

[EDIT]However, you did not state what the field axioms are. Do they allow all the steps that you made?[/EDIT]

 

[HallsofIvy]

You say that you are using the field axioms, one of which is that every member of the field except 0 has a multiplicative inverse: If ab= ac and a is not 0, then $a^{-1}(ab)= (a^{-1}a)b$ (by the associative law) so $a^{-1}(ab)= 1(b)= b$ while [tex]a^{-1}(ac)= (a^{-1}a)c= 1(c)= c[/itex]. That is, multiplying both sides of ab= ac by $a^{-1}$ gives b= c.<br /> <br /> A more general object is the "cancelation ring" in which non-zero members do NOT necessarily have multiplicative inverses but if ab= 0 and a is not 0, then b= 0. In that case, if ab= ac, ab- ac= a(b- c)= 0. Since a is not 0, we must have b- c= 0 and so b= c. An example of a cancelation ring that is not a field is the set of all integers with ordinary addition and multiplication.[/tex]

 

[gerimis]

ab=ac
It means
10a+b =10a +c
b=c

---
If you don't understand why ab=10a+b,

It will be easier to explain If i give an example:

45=10*(4)+5
ab=10*(a)+b

 

[Robert1986]

ab=ac
It means
10a+b =10a +c
b=c

---
If you don't understand why ab=10a+b,

It will be easier to explain If i give an example:

45=10*(4)+5
ab=10*(a)+b

You don't seem to understand what we're doing, here.

The OP is referring to a general Field, and when he writes ab he means a*b where * is the multiplication in the Field. He does not mean what you seem to think he means. That is, ab is not a number written in decimal notation.

 

[kripkrip420]

There is nothing wrong with your proof. The assumption that a <> 0 needs to hold to show b = c. If a was zero then a solution b <> c would exist which is why they stated that a must be non-zero.

Thank you for the clarification and your reply!

 

[kripkrip420]

You say that you are using the field axioms, one of which is that every member of the field except 0 has a multiplicative inverse: If ab= ac and a is not 0, then $a^{-1}(ab)= (a^{-1}a)b$ (by the associative law) so $a^{-1}(ab)= 1(b)= b$ while [tex]a^{-1}(ac)= (a^{-1}a)c= 1(c)= c[/itex]. That is, multiplying both sides of ab= ac by $a^{-1}$ gives b= c.<br /> <br /> A more general object is the "cancelation ring" in which non-zero members do NOT necessarily have multiplicative inverses but if ab= 0 and a is not 0, then b= 0. In that case, if ab= ac, ab- ac= a(b- c)= 0. Since a is not 0, we must have b- c= 0 and so b= c. An example of a cancelation ring that is not a field is the set of all integers with ordinary addition and multiplication.[/tex]

$$<br /> Sorry. I have forgotten to include that no division axioms may be used but nonetheless, thank you!$$

 

[kripkrip420]

Thanks to all who replied! You have helped me greatly!