Mixing
Is this proof of the Cancellation Law for multiplication of real numbers correct?
Clash Royale CLAN TAG#URR8PPP
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The original statement is:
If $ab = ac$ and $a neq 0$, then $b = c$. (In particular, this shows that the number 1 of Axiom 4 is unique.)
By Axiom 6, we know that exists a real number y such that $y*a = 1$. Since multiplications are uniquely defined, we know that $y*(a*b) = y*(a*c)$. By the Axiom 2, we can rewrite it as $(y*a)*b = (y*a)*c$ but we do know that $y*a=1$ so we have $1*b = 1*c$. By Axiom 4, we know that $1*b = b$ and $1*c =c$ so we have $b = c$
Axiom 2: $x + (y + z) = (x + y) + z$ and $x(yz) = (xy)z$.
Axiom 4: There exist two distinct real numbers, which
we denote by 0 and 1, such that for every real x we have $x + 0 = x$ and $1*x = x$.
Axiom 6: For every real number $x neq 0$ there is a real
number y such that $x*y = 1$.
proof-verification real-numbers
edited Jul 21 at 1:42
Mike Pierce
11k93574
asked Jul 21 at 1:14
Victor Feitosa
445
add a comment |Â
up vote
0
down vote
favorite
The original statement is:
If $ab = ac$ and $a neq 0$, then $b = c$. (In particular, this shows that the number 1 of Axiom 4 is unique.)
By Axiom 6, we know that exists a real number y such that $y*a = 1$. Since multiplications are uniquely defined, we know that $y*(a*b) = y*(a*c)$. By the Axiom 2, we can rewrite it as $(y*a)*b = (y*a)*c$ but we do know that $y*a=1$ so we have $1*b = 1*c$. By Axiom 4, we know that $1*b = b$ and $1*c =c$ so we have $b = c$
Axiom 2: $x + (y + z) = (x + y) + z$ and $x(yz) = (xy)z$.
Axiom 4: There exist two distinct real numbers, which
we denote by 0 and 1, such that for every real x we have $x + 0 = x$ and $1*x = x$.
Axiom 6: For every real number $x neq 0$ there is a real
number y such that $x*y = 1$.
proof-verification real-numbers
edited Jul 21 at 1:42
Mike Pierce
11k93574
asked Jul 21 at 1:14
Victor Feitosa
445
1
It's perfect...............
– DanielWainfleet
Jul 21 at 1:25
" Since multiplications are uniquely defined" I'm not entirely sure what you mean by that and that's probably not the best wording for what you are trying to say. But otherwise your proof is absolutely correct and quite well done.
– fleablood
Jul 21 at 5:17
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The original statement is:
If $ab = ac$ and $a neq 0$, then $b = c$. (In particular, this shows that the number 1 of Axiom 4 is unique.)
By Axiom 6, we know that exists a real number y such that $y*a = 1$. Since multiplications are uniquely defined, we know that $y*(a*b) = y*(a*c)$. By the Axiom 2, we can rewrite it as $(y*a)*b = (y*a)*c$ but we do know that $y*a=1$ so we have $1*b = 1*c$. By Axiom 4, we know that $1*b = b$ and $1*c =c$ so we have $b = c$
Axiom 2: $x + (y + z) = (x + y) + z$ and $x(yz) = (xy)z$.
Axiom 4: There exist two distinct real numbers, which
we denote by 0 and 1, such that for every real x we have $x + 0 = x$ and $1*x = x$.
Axiom 6: For every real number $x neq 0$ there is a real
number y such that $x*y = 1$.
proof-verification real-numbers
edited Jul 21 at 1:42
Mike Pierce
11k93574
asked Jul 21 at 1:14
Victor Feitosa
445
The original statement is:
If $ab = ac$ and $a neq 0$, then $b = c$. (In particular, this shows that the number 1 of Axiom 4 is unique.)
By Axiom 6, we know that exists a real number y such that $y*a = 1$. Since multiplications are uniquely defined, we know that $y*(a*b) = y*(a*c)$. By the Axiom 2, we can rewrite it as $(y*a)*b = (y*a)*c$ but we do know that $y*a=1$ so we have $1*b = 1*c$. By Axiom 4, we know that $1*b = b$ and $1*c =c$ so we have $b = c$
Axiom 2: $x + (y + z) = (x + y) + z$ and $x(yz) = (xy)z$.
Axiom 4: There exist two distinct real numbers, which
we denote by 0 and 1, such that for every real x we have $x + 0 = x$ and $1*x = x$.
Axiom 6: For every real number $x neq 0$ there is a real
number y such that $x*y = 1$.
proof-verification real-numbers
edited Jul 21 at 1:42
Mike Pierce
11k93574
asked Jul 21 at 1:14
Victor Feitosa
445
edited Jul 21 at 1:42
Mike Pierce
11k93574
edited Jul 21 at 1:42
Mike Pierce
11k93574
edited Jul 21 at 1:42
Mike Pierce
11k93574
11k93574
asked Jul 21 at 1:14
Victor Feitosa
445
asked Jul 21 at 1:14
Victor Feitosa
445
asked Jul 21 at 1:14
Victor Feitosa
445
445
1
It's perfect...............
– DanielWainfleet
Jul 21 at 1:25
" Since multiplications are uniquely defined" I'm not entirely sure what you mean by that and that's probably not the best wording for what you are trying to say. But otherwise your proof is absolutely correct and quite well done.
– fleablood
Jul 21 at 5:17
add a comment |Â
1
It's perfect...............
– DanielWainfleet
Jul 21 at 1:25
" Since multiplications are uniquely defined" I'm not entirely sure what you mean by that and that's probably not the best wording for what you are trying to say. But otherwise your proof is absolutely correct and quite well done.
– fleablood
Jul 21 at 5:17
1
1
It's perfect...............
– DanielWainfleet
Jul 21 at 1:25
It's perfect...............
– DanielWainfleet
Jul 21 at 1:25
" Since multiplications are uniquely defined" I'm not entirely sure what you mean by that and that's probably not the best wording for what you are trying to say. But otherwise your proof is absolutely correct and quite well done.
– fleablood
Jul 21 at 5:17
" Since multiplications are uniquely defined" I'm not entirely sure what you mean by that and that's probably not the best wording for what you are trying to say. But otherwise your proof is absolutely correct and quite well done.
– fleablood
Jul 21 at 5:17
add a comment |Â
1 Answer
1
active
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up vote
5
down vote
accepted
Steven Stadnicki's is correct that your proof looks fine.
I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue.
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Steven Stadnicki's is correct that your proof looks fine.
I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue.
add a comment |Â
up vote
5
down vote
accepted
Steven Stadnicki's is correct that your proof looks fine.
I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue.
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Steven Stadnicki's is correct that your proof looks fine.
I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue.
Steven Stadnicki's is correct that your proof looks fine.
I'm posting this CW answer so that users who confidently concur have something to vote on, and so this question doesn't stagnate in the Unanswered Questions Queue.
answered Jul 21 at 1:40
community wiki
Mike Pierce
add a comment |Â
add a comment |Â
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1
It's perfect...............
– DanielWainfleet
Jul 21 at 1:25
" Since multiplications are uniquely defined" I'm not entirely sure what you mean by that and that's probably not the best wording for what you are trying to say. But otherwise your proof is absolutely correct and quite well done.
– fleablood
Jul 21 at 5:17