Mixing
Is this proof of $0 * a = 0$ correct?
Clash Royale CLAN TAG#URR8PPP
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I need to prove that $0*a = a*0 = 0$ using only axiomatic properties for real numbers(It's a exercise from Apostol's book.
My try:
We can write $0$ as $(a - a)$. Hence we can rewrite the equation as follows:
$0*a = a*0 = a*(a-a) = 0$
By distributive law we have: $aa -aa = 0$, where $aa - aa$ evaluates to $0$ because $(-aa)$ is the negative of $aa$
Note: It's given that $a(b - c) = ab - ac.$ (Theorem 1.4)
proof-verification
edited Jul 18 at 1:01
RayDansh
884215
asked Jul 18 at 0:46
Victor Feitosa
445
add a comment |Â
up vote
2
down vote
favorite
I need to prove that $0*a = a*0 = 0$ using only axiomatic properties for real numbers(It's a exercise from Apostol's book.
My try:
We can write $0$ as $(a - a)$. Hence we can rewrite the equation as follows:
$0*a = a*0 = a*(a-a) = 0$
By distributive law we have: $aa -aa = 0$, where $aa - aa$ evaluates to $0$ because $(-aa)$ is the negative of $aa$
Note: It's given that $a(b - c) = ab - ac.$ (Theorem 1.4)
proof-verification
edited Jul 18 at 1:01
RayDansh
884215
asked Jul 18 at 0:46
Victor Feitosa
445
1
If you have already proven that the additive inverse of $aa$ is $-aa$ then your proof works great. If you haven't proven that yet, then you probably shouldn't assume it is true.
– InterstellarProbe
Jul 18 at 0:57
1
An alternative is to use $0=0+0$, multiply by $a$, use the distributive law, and inverses.
– Michael Burr
Jul 18 at 1:09
I forgot to mention but it's also given that the additive inverse of aa is -aa. Thanks! I'll try to proof using it @MichaelBurr
– Victor Feitosa
Jul 18 at 1:21
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to prove that $0*a = a*0 = 0$ using only axiomatic properties for real numbers(It's a exercise from Apostol's book.
My try:
We can write $0$ as $(a - a)$. Hence we can rewrite the equation as follows:
$0*a = a*0 = a*(a-a) = 0$
By distributive law we have: $aa -aa = 0$, where $aa - aa$ evaluates to $0$ because $(-aa)$ is the negative of $aa$
Note: It's given that $a(b - c) = ab - ac.$ (Theorem 1.4)
proof-verification
edited Jul 18 at 1:01
RayDansh
884215
asked Jul 18 at 0:46
Victor Feitosa
445
I need to prove that $0*a = a*0 = 0$ using only axiomatic properties for real numbers(It's a exercise from Apostol's book.
My try:
We can write $0$ as $(a - a)$. Hence we can rewrite the equation as follows:
$0*a = a*0 = a*(a-a) = 0$
By distributive law we have: $aa -aa = 0$, where $aa - aa$ evaluates to $0$ because $(-aa)$ is the negative of $aa$
Note: It's given that $a(b - c) = ab - ac.$ (Theorem 1.4)
proof-verification
edited Jul 18 at 1:01
RayDansh
884215
asked Jul 18 at 0:46
Victor Feitosa
445
edited Jul 18 at 1:01
RayDansh
884215
edited Jul 18 at 1:01
RayDansh
884215
edited Jul 18 at 1:01
RayDansh
884215
884215
asked Jul 18 at 0:46
Victor Feitosa
445
asked Jul 18 at 0:46
Victor Feitosa
445
asked Jul 18 at 0:46
Victor Feitosa
445
445
1
If you have already proven that the additive inverse of $aa$ is $-aa$ then your proof works great. If you haven't proven that yet, then you probably shouldn't assume it is true.
– InterstellarProbe
Jul 18 at 0:57
1
An alternative is to use $0=0+0$, multiply by $a$, use the distributive law, and inverses.
– Michael Burr
Jul 18 at 1:09
I forgot to mention but it's also given that the additive inverse of aa is -aa. Thanks! I'll try to proof using it @MichaelBurr
– Victor Feitosa
Jul 18 at 1:21
add a comment |Â
1
If you have already proven that the additive inverse of $aa$ is $-aa$ then your proof works great. If you haven't proven that yet, then you probably shouldn't assume it is true.
– InterstellarProbe
Jul 18 at 0:57
1
An alternative is to use $0=0+0$, multiply by $a$, use the distributive law, and inverses.
– Michael Burr
Jul 18 at 1:09
I forgot to mention but it's also given that the additive inverse of aa is -aa. Thanks! I'll try to proof using it @MichaelBurr
– Victor Feitosa
Jul 18 at 1:21
1
1
If you have already proven that the additive inverse of $aa$ is $-aa$ then your proof works great. If you haven't proven that yet, then you probably shouldn't assume it is true.
– InterstellarProbe
Jul 18 at 0:57
If you have already proven that the additive inverse of $aa$ is $-aa$ then your proof works great. If you haven't proven that yet, then you probably shouldn't assume it is true.
– InterstellarProbe
Jul 18 at 0:57
1
1
An alternative is to use $0=0+0$, multiply by $a$, use the distributive law, and inverses.
– Michael Burr
Jul 18 at 1:09
An alternative is to use $0=0+0$, multiply by $a$, use the distributive law, and inverses.
– Michael Burr
Jul 18 at 1:09
I forgot to mention but it's also given that the additive inverse of aa is -aa. Thanks! I'll try to proof using it @MichaelBurr
– Victor Feitosa
Jul 18 at 1:21
I forgot to mention but it's also given that the additive inverse of aa is -aa. Thanks! I'll try to proof using it @MichaelBurr
– Victor Feitosa
Jul 18 at 1:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Given the Theorem 1.4, this proof seems fine. It uses only the results available to you and the axioms.
Unfortunately, there is something bugging me in general about the proof. One has to be careful with establishing an arithmetic with minus signs. Little things can trip you up. For example, it's natural to look at $(-a)b$ and think it just means $-(ab)$, but conceptually, they're different.
Yes, the theorem does seem to cover this, but it still bothers me, because establishing Theorem 1.4 often relies on establishing $0a = 0$ first. Here's the way I'm used to the proof going:
$$0a = (0 + 0)a = 0a + 0a,$$
which implies $0a = 0$ by cancellation (add the inverse of $0a$ to both sides, and simplify). Then, to establish that $(-a)b = -(ab) = a(-b)$, we observe that
$$(-a)b + ab = ((-a) + a)b = 0b = 0,$$
and similarly $ab + (-a)b = 0$. Therefore $(-a)b$ fits the definition of the inverse of $ab$. Similarly, $a(-b) = -(ab)$ as well. Finally, this yields Theorem 1.4:
$$ab - ac = ab + (-(ac)) = ab + a(-c) = a(b + (-c)) = a(b - c).$$
I don't really know how you'd prove Theorem 1.4 without relying on $0a = 0$.
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
Maybe it helps, but it's first proven that $a - b = a + (-b)$
– Victor Feitosa
Jul 18 at 1:16
@VictorFeitosa Well, to me, that's the very definition of $a - b$: the sum of $a$ and the additive inverse of $b$. I use that in the final stage of the proof.
– Theo Bendit
Jul 18 at 1:18
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Given the Theorem 1.4, this proof seems fine. It uses only the results available to you and the axioms.
Unfortunately, there is something bugging me in general about the proof. One has to be careful with establishing an arithmetic with minus signs. Little things can trip you up. For example, it's natural to look at $(-a)b$ and think it just means $-(ab)$, but conceptually, they're different.
Yes, the theorem does seem to cover this, but it still bothers me, because establishing Theorem 1.4 often relies on establishing $0a = 0$ first. Here's the way I'm used to the proof going:
$$0a = (0 + 0)a = 0a + 0a,$$
which implies $0a = 0$ by cancellation (add the inverse of $0a$ to both sides, and simplify). Then, to establish that $(-a)b = -(ab) = a(-b)$, we observe that
$$(-a)b + ab = ((-a) + a)b = 0b = 0,$$
and similarly $ab + (-a)b = 0$. Therefore $(-a)b$ fits the definition of the inverse of $ab$. Similarly, $a(-b) = -(ab)$ as well. Finally, this yields Theorem 1.4:
$$ab - ac = ab + (-(ac)) = ab + a(-c) = a(b + (-c)) = a(b - c).$$
I don't really know how you'd prove Theorem 1.4 without relying on $0a = 0$.
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
Maybe it helps, but it's first proven that $a - b = a + (-b)$
– Victor Feitosa
Jul 18 at 1:16
@VictorFeitosa Well, to me, that's the very definition of $a - b$: the sum of $a$ and the additive inverse of $b$. I use that in the final stage of the proof.
– Theo Bendit
Jul 18 at 1:18
add a comment |Â
up vote
2
down vote
accepted
Given the Theorem 1.4, this proof seems fine. It uses only the results available to you and the axioms.
Unfortunately, there is something bugging me in general about the proof. One has to be careful with establishing an arithmetic with minus signs. Little things can trip you up. For example, it's natural to look at $(-a)b$ and think it just means $-(ab)$, but conceptually, they're different.
Yes, the theorem does seem to cover this, but it still bothers me, because establishing Theorem 1.4 often relies on establishing $0a = 0$ first. Here's the way I'm used to the proof going:
$$0a = (0 + 0)a = 0a + 0a,$$
which implies $0a = 0$ by cancellation (add the inverse of $0a$ to both sides, and simplify). Then, to establish that $(-a)b = -(ab) = a(-b)$, we observe that
$$(-a)b + ab = ((-a) + a)b = 0b = 0,$$
and similarly $ab + (-a)b = 0$. Therefore $(-a)b$ fits the definition of the inverse of $ab$. Similarly, $a(-b) = -(ab)$ as well. Finally, this yields Theorem 1.4:
$$ab - ac = ab + (-(ac)) = ab + a(-c) = a(b + (-c)) = a(b - c).$$
I don't really know how you'd prove Theorem 1.4 without relying on $0a = 0$.
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
Maybe it helps, but it's first proven that $a - b = a + (-b)$
– Victor Feitosa
Jul 18 at 1:16
@VictorFeitosa Well, to me, that's the very definition of $a - b$: the sum of $a$ and the additive inverse of $b$. I use that in the final stage of the proof.
– Theo Bendit
Jul 18 at 1:18
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Given the Theorem 1.4, this proof seems fine. It uses only the results available to you and the axioms.
Unfortunately, there is something bugging me in general about the proof. One has to be careful with establishing an arithmetic with minus signs. Little things can trip you up. For example, it's natural to look at $(-a)b$ and think it just means $-(ab)$, but conceptually, they're different.
Yes, the theorem does seem to cover this, but it still bothers me, because establishing Theorem 1.4 often relies on establishing $0a = 0$ first. Here's the way I'm used to the proof going:
$$0a = (0 + 0)a = 0a + 0a,$$
which implies $0a = 0$ by cancellation (add the inverse of $0a$ to both sides, and simplify). Then, to establish that $(-a)b = -(ab) = a(-b)$, we observe that
$$(-a)b + ab = ((-a) + a)b = 0b = 0,$$
and similarly $ab + (-a)b = 0$. Therefore $(-a)b$ fits the definition of the inverse of $ab$. Similarly, $a(-b) = -(ab)$ as well. Finally, this yields Theorem 1.4:
$$ab - ac = ab + (-(ac)) = ab + a(-c) = a(b + (-c)) = a(b - c).$$
I don't really know how you'd prove Theorem 1.4 without relying on $0a = 0$.
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
Given the Theorem 1.4, this proof seems fine. It uses only the results available to you and the axioms.
Unfortunately, there is something bugging me in general about the proof. One has to be careful with establishing an arithmetic with minus signs. Little things can trip you up. For example, it's natural to look at $(-a)b$ and think it just means $-(ab)$, but conceptually, they're different.
Yes, the theorem does seem to cover this, but it still bothers me, because establishing Theorem 1.4 often relies on establishing $0a = 0$ first. Here's the way I'm used to the proof going:
$$0a = (0 + 0)a = 0a + 0a,$$
which implies $0a = 0$ by cancellation (add the inverse of $0a$ to both sides, and simplify). Then, to establish that $(-a)b = -(ab) = a(-b)$, we observe that
$$(-a)b + ab = ((-a) + a)b = 0b = 0,$$
and similarly $ab + (-a)b = 0$. Therefore $(-a)b$ fits the definition of the inverse of $ab$. Similarly, $a(-b) = -(ab)$ as well. Finally, this yields Theorem 1.4:
$$ab - ac = ab + (-(ac)) = ab + a(-c) = a(b + (-c)) = a(b - c).$$
I don't really know how you'd prove Theorem 1.4 without relying on $0a = 0$.
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
answered Jul 18 at 1:13
Theo Bendit
12.1k1844
12.1k1844
Maybe it helps, but it's first proven that $a - b = a + (-b)$
– Victor Feitosa
Jul 18 at 1:16
@VictorFeitosa Well, to me, that's the very definition of $a - b$: the sum of $a$ and the additive inverse of $b$. I use that in the final stage of the proof.
– Theo Bendit
Jul 18 at 1:18
add a comment |Â
Maybe it helps, but it's first proven that $a - b = a + (-b)$
– Victor Feitosa
Jul 18 at 1:16
@VictorFeitosa Well, to me, that's the very definition of $a - b$: the sum of $a$ and the additive inverse of $b$. I use that in the final stage of the proof.
– Theo Bendit
Jul 18 at 1:18
Maybe it helps, but it's first proven that $a - b = a + (-b)$
– Victor Feitosa
Jul 18 at 1:16
Maybe it helps, but it's first proven that $a - b = a + (-b)$
– Victor Feitosa
Jul 18 at 1:16
@VictorFeitosa Well, to me, that's the very definition of $a - b$: the sum of $a$ and the additive inverse of $b$. I use that in the final stage of the proof.
– Theo Bendit
Jul 18 at 1:18
@VictorFeitosa Well, to me, that's the very definition of $a - b$: the sum of $a$ and the additive inverse of $b$. I use that in the final stage of the proof.
– Theo Bendit
Jul 18 at 1:18
add a comment |Â
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1
If you have already proven that the additive inverse of $aa$ is $-aa$ then your proof works great. If you haven't proven that yet, then you probably shouldn't assume it is true.
– InterstellarProbe
Jul 18 at 0:57
1
An alternative is to use $0=0+0$, multiply by $a$, use the distributive law, and inverses.
– Michael Burr
Jul 18 at 1:09
I forgot to mention but it's also given that the additive inverse of aa is -aa. Thanks! I'll try to proof using it @MichaelBurr
– Victor Feitosa
Jul 18 at 1:21