Mixing
Deductions using Field axioms
$begingroup$
Deduce $alpha(-beta)=(-alpha)(beta)=-(alphabeta)$
I'm struggling with where to start and which axioms will be useful.
The axioms:
Axioms for a field $mathbb{K}$ are with respect to the additive operation $+ : mathbb{K} times mathbb{K} rightarrow mathbb{K}$:
1) $forall x,y,zin mathbb{K}: (x+y)+z = x+(y+z)$
2) $forall x,y in mathbb{K}: x+y = y+x$
3) $exists 0 in mathbb{K} $such that$ : x+0=x$
4) $forall x in mathbb{K} exists y in mathbb{K}: x+y=0 rightarrow y = -x$
for the multiplicative operation:
1) $forall x,y,z in mathbb{K}: (xy)z = x(yz)$
2) $forall x,y in mathbb{K}: xy=yx$
3) $exists 1 in K : forall x in mathbb{K}: x1=x$
4) $forall xneq 0 in mathbb{K} exists y in mathbb{K}: xy=1 rightarrow y = x^{-1}= frac{1}{x}$
and the distributivity property
$(x+y)z=xz+yz$
abstract-algebra field-theory
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
asked Jan 27 at 17:04
m.bazzam.bazza
948
$endgroup$
add a comment |
$begingroup$
Deduce $alpha(-beta)=(-alpha)(beta)=-(alphabeta)$
I'm struggling with where to start and which axioms will be useful.
The axioms:
Axioms for a field $mathbb{K}$ are with respect to the additive operation $+ : mathbb{K} times mathbb{K} rightarrow mathbb{K}$:
1) $forall x,y,zin mathbb{K}: (x+y)+z = x+(y+z)$
2) $forall x,y in mathbb{K}: x+y = y+x$
3) $exists 0 in mathbb{K} $such that$ : x+0=x$
4) $forall x in mathbb{K} exists y in mathbb{K}: x+y=0 rightarrow y = -x$
for the multiplicative operation:
1) $forall x,y,z in mathbb{K}: (xy)z = x(yz)$
2) $forall x,y in mathbb{K}: xy=yx$
3) $exists 1 in K : forall x in mathbb{K}: x1=x$
4) $forall xneq 0 in mathbb{K} exists y in mathbb{K}: xy=1 rightarrow y = x^{-1}= frac{1}{x}$
and the distributivity property
$(x+y)z=xz+yz$
abstract-algebra field-theory
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
asked Jan 27 at 17:04
m.bazzam.bazza
948
$endgroup$
$begingroup$
Possibly factoring out your "$-1$" through commutativity and associativity?
$endgroup$
– Hyperion
Jan 27 at 17:08
$begingroup$
You'll increase your chances of getting an answer if you include the axioms. As for a place to start, note the following: for any $x$ in the given field, $x=-(alphabeta) iff x+alphabeta=0$, this is by definition of additive inverse.
$endgroup$
– Git Gud
Jan 27 at 17:09
add a comment |
$begingroup$
Deduce $alpha(-beta)=(-alpha)(beta)=-(alphabeta)$
I'm struggling with where to start and which axioms will be useful.
The axioms:
Axioms for a field $mathbb{K}$ are with respect to the additive operation $+ : mathbb{K} times mathbb{K} rightarrow mathbb{K}$:
1) $forall x,y,zin mathbb{K}: (x+y)+z = x+(y+z)$
2) $forall x,y in mathbb{K}: x+y = y+x$
3) $exists 0 in mathbb{K} $such that$ : x+0=x$
4) $forall x in mathbb{K} exists y in mathbb{K}: x+y=0 rightarrow y = -x$
for the multiplicative operation:
1) $forall x,y,z in mathbb{K}: (xy)z = x(yz)$
2) $forall x,y in mathbb{K}: xy=yx$
3) $exists 1 in K : forall x in mathbb{K}: x1=x$
4) $forall xneq 0 in mathbb{K} exists y in mathbb{K}: xy=1 rightarrow y = x^{-1}= frac{1}{x}$
and the distributivity property
$(x+y)z=xz+yz$
abstract-algebra field-theory
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
asked Jan 27 at 17:04
m.bazzam.bazza
948
$endgroup$
Deduce $alpha(-beta)=(-alpha)(beta)=-(alphabeta)$
I'm struggling with where to start and which axioms will be useful.
The axioms:
Axioms for a field $mathbb{K}$ are with respect to the additive operation $+ : mathbb{K} times mathbb{K} rightarrow mathbb{K}$:
1) $forall x,y,zin mathbb{K}: (x+y)+z = x+(y+z)$
2) $forall x,y in mathbb{K}: x+y = y+x$
3) $exists 0 in mathbb{K} $such that$ : x+0=x$
4) $forall x in mathbb{K} exists y in mathbb{K}: x+y=0 rightarrow y = -x$
for the multiplicative operation:
1) $forall x,y,z in mathbb{K}: (xy)z = x(yz)$
2) $forall x,y in mathbb{K}: xy=yx$
3) $exists 1 in K : forall x in mathbb{K}: x1=x$
4) $forall xneq 0 in mathbb{K} exists y in mathbb{K}: xy=1 rightarrow y = x^{-1}= frac{1}{x}$
and the distributivity property
$(x+y)z=xz+yz$
abstract-algebra field-theory
abstract-algebra field-theory
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
asked Jan 27 at 17:04
m.bazzam.bazza
948
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
asked Jan 27 at 17:04
m.bazzam.bazza
948
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
edited Jan 27 at 18:19
J. W. Tanner
3,8251320
3,8251320
asked Jan 27 at 17:04
m.bazzam.bazza
948
asked Jan 27 at 17:04
m.bazzam.bazza
948
asked Jan 27 at 17:04
m.bazzam.bazza
948
948
$begingroup$
Possibly factoring out your "$-1$" through commutativity and associativity?
$endgroup$
– Hyperion
Jan 27 at 17:08
$begingroup$
You'll increase your chances of getting an answer if you include the axioms. As for a place to start, note the following: for any $x$ in the given field, $x=-(alphabeta) iff x+alphabeta=0$, this is by definition of additive inverse.
$endgroup$
– Git Gud
Jan 27 at 17:09
add a comment |
$begingroup$
Possibly factoring out your "$-1$" through commutativity and associativity?
$endgroup$
– Hyperion
Jan 27 at 17:08
$begingroup$
You'll increase your chances of getting an answer if you include the axioms. As for a place to start, note the following: for any $x$ in the given field, $x=-(alphabeta) iff x+alphabeta=0$, this is by definition of additive inverse.
$endgroup$
– Git Gud
Jan 27 at 17:09
$begingroup$
Possibly factoring out your "$-1$" through commutativity and associativity?
$endgroup$
– Hyperion
Jan 27 at 17:08
$begingroup$
Possibly factoring out your "$-1$" through commutativity and associativity?
$endgroup$
– Hyperion
Jan 27 at 17:08
$begingroup$
You'll increase your chances of getting an answer if you include the axioms. As for a place to start, note the following: for any $x$ in the given field, $x=-(alphabeta) iff x+alphabeta=0$, this is by definition of additive inverse.
$endgroup$
– Git Gud
Jan 27 at 17:09
$begingroup$
You'll increase your chances of getting an answer if you include the axioms. As for a place to start, note the following: for any $x$ in the given field, $x=-(alphabeta) iff x+alphabeta=0$, this is by definition of additive inverse.
$endgroup$
– Git Gud
Jan 27 at 17:09
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You'll have to first prove a lemma using the axioms, namely that $x cdot 0 = 0 cdot x = 0$ for all $x$.
Once you've done that, then
$$alphabeta + alpha(-beta) = alpha(beta + (-beta)) = alpha cdot 0 = 0
$$
and so $alpha(-beta)$ is equal to the additive inverse of $alphabeta$ which is $-alphabeta$.
Similarly for $(-alpha)beta$.
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
$endgroup$
add a comment |
$begingroup$
Attempt at solving equality $(alpha)(-beta)=-(alphabeta)$
$alpha(-beta)=alpha((-1)(beta)=(alpha(-1))(beta)=((-1)alpha)beta)=(-1)((alpha)(beta))=(-1)(alphabeta)=-(alphabeta)$
answered Jan 27 at 17:20
m.bazzam.bazza
948
$endgroup$
add a comment |
$begingroup$
Hint:
Use this fact: For all $alpha$, $-alpha = (-1)cdot alpha$.
Proof of the fact: $alpha+(-1)cdot alpha = 1cdot alpha + (-1)cdotalpha = (1-1)cdot alpha = 0cdot alpha = 0$.
The last equality holds because $0cdot alpha = (0+0)cdot alpha = 0cdotalpha + 0 cdot alpha$ and so by removing $0cdot alpha$ from both sides we get $0cdotalpha = 0$.
From this fact you can deduce all of the equalities. I'll do one of them and leave the rest for you
Example: $alpha cdot (-beta) = (-alpha)beta$
Proof: $alphacdot (-beta) = alphacdot ((-1)cdot beta) = (alpha cdot (-1))cdot beta = ((-1)cdot alpha)cdot beta = (-alpha)cdot beta$.
answered Jan 27 at 17:09
YankoYanko
8,0832830
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You'll have to first prove a lemma using the axioms, namely that $x cdot 0 = 0 cdot x = 0$ for all $x$.
Once you've done that, then
$$alphabeta + alpha(-beta) = alpha(beta + (-beta)) = alpha cdot 0 = 0
$$
and so $alpha(-beta)$ is equal to the additive inverse of $alphabeta$ which is $-alphabeta$.
Similarly for $(-alpha)beta$.
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
$endgroup$
add a comment |
$begingroup$
You'll have to first prove a lemma using the axioms, namely that $x cdot 0 = 0 cdot x = 0$ for all $x$.
Once you've done that, then
$$alphabeta + alpha(-beta) = alpha(beta + (-beta)) = alpha cdot 0 = 0
$$
and so $alpha(-beta)$ is equal to the additive inverse of $alphabeta$ which is $-alphabeta$.
Similarly for $(-alpha)beta$.
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
$endgroup$
add a comment |
$begingroup$
You'll have to first prove a lemma using the axioms, namely that $x cdot 0 = 0 cdot x = 0$ for all $x$.
Once you've done that, then
$$alphabeta + alpha(-beta) = alpha(beta + (-beta)) = alpha cdot 0 = 0
$$
and so $alpha(-beta)$ is equal to the additive inverse of $alphabeta$ which is $-alphabeta$.
Similarly for $(-alpha)beta$.
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
$endgroup$
You'll have to first prove a lemma using the axioms, namely that $x cdot 0 = 0 cdot x = 0$ for all $x$.
Once you've done that, then
$$alphabeta + alpha(-beta) = alpha(beta + (-beta)) = alpha cdot 0 = 0
$$
and so $alpha(-beta)$ is equal to the additive inverse of $alphabeta$ which is $-alphabeta$.
Similarly for $(-alpha)beta$.
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
answered Jan 27 at 18:43
Lee MosherLee Mosher
51.2k33889
51.2k33889
add a comment |
add a comment |
$begingroup$
Attempt at solving equality $(alpha)(-beta)=-(alphabeta)$
$alpha(-beta)=alpha((-1)(beta)=(alpha(-1))(beta)=((-1)alpha)beta)=(-1)((alpha)(beta))=(-1)(alphabeta)=-(alphabeta)$
answered Jan 27 at 17:20
m.bazzam.bazza
948
$endgroup$
add a comment |
$begingroup$
Attempt at solving equality $(alpha)(-beta)=-(alphabeta)$
$alpha(-beta)=alpha((-1)(beta)=(alpha(-1))(beta)=((-1)alpha)beta)=(-1)((alpha)(beta))=(-1)(alphabeta)=-(alphabeta)$
answered Jan 27 at 17:20
m.bazzam.bazza
948
$endgroup$
add a comment |
$begingroup$
Attempt at solving equality $(alpha)(-beta)=-(alphabeta)$
$alpha(-beta)=alpha((-1)(beta)=(alpha(-1))(beta)=((-1)alpha)beta)=(-1)((alpha)(beta))=(-1)(alphabeta)=-(alphabeta)$
answered Jan 27 at 17:20
m.bazzam.bazza
948
$endgroup$
Attempt at solving equality $(alpha)(-beta)=-(alphabeta)$
$alpha(-beta)=alpha((-1)(beta)=(alpha(-1))(beta)=((-1)alpha)beta)=(-1)((alpha)(beta))=(-1)(alphabeta)=-(alphabeta)$
answered Jan 27 at 17:20
m.bazzam.bazza
948
answered Jan 27 at 17:20
m.bazzam.bazza
948
answered Jan 27 at 17:20
m.bazzam.bazza
948
answered Jan 27 at 17:20
m.bazzam.bazza
948
948
add a comment |
add a comment |
$begingroup$
Hint:
Use this fact: For all $alpha$, $-alpha = (-1)cdot alpha$.
Proof of the fact: $alpha+(-1)cdot alpha = 1cdot alpha + (-1)cdotalpha = (1-1)cdot alpha = 0cdot alpha = 0$.
The last equality holds because $0cdot alpha = (0+0)cdot alpha = 0cdotalpha + 0 cdot alpha$ and so by removing $0cdot alpha$ from both sides we get $0cdotalpha = 0$.
From this fact you can deduce all of the equalities. I'll do one of them and leave the rest for you
Example: $alpha cdot (-beta) = (-alpha)beta$
Proof: $alphacdot (-beta) = alphacdot ((-1)cdot beta) = (alpha cdot (-1))cdot beta = ((-1)cdot alpha)cdot beta = (-alpha)cdot beta$.
answered Jan 27 at 17:09
YankoYanko
8,0832830
$endgroup$
add a comment |
$begingroup$
Hint:
Use this fact: For all $alpha$, $-alpha = (-1)cdot alpha$.
Proof of the fact: $alpha+(-1)cdot alpha = 1cdot alpha + (-1)cdotalpha = (1-1)cdot alpha = 0cdot alpha = 0$.
The last equality holds because $0cdot alpha = (0+0)cdot alpha = 0cdotalpha + 0 cdot alpha$ and so by removing $0cdot alpha$ from both sides we get $0cdotalpha = 0$.
From this fact you can deduce all of the equalities. I'll do one of them and leave the rest for you
Example: $alpha cdot (-beta) = (-alpha)beta$
Proof: $alphacdot (-beta) = alphacdot ((-1)cdot beta) = (alpha cdot (-1))cdot beta = ((-1)cdot alpha)cdot beta = (-alpha)cdot beta$.
answered Jan 27 at 17:09
YankoYanko
8,0832830
$endgroup$
add a comment |
$begingroup$
Hint:
Use this fact: For all $alpha$, $-alpha = (-1)cdot alpha$.
Proof of the fact: $alpha+(-1)cdot alpha = 1cdot alpha + (-1)cdotalpha = (1-1)cdot alpha = 0cdot alpha = 0$.
The last equality holds because $0cdot alpha = (0+0)cdot alpha = 0cdotalpha + 0 cdot alpha$ and so by removing $0cdot alpha$ from both sides we get $0cdotalpha = 0$.
From this fact you can deduce all of the equalities. I'll do one of them and leave the rest for you
Example: $alpha cdot (-beta) = (-alpha)beta$
Proof: $alphacdot (-beta) = alphacdot ((-1)cdot beta) = (alpha cdot (-1))cdot beta = ((-1)cdot alpha)cdot beta = (-alpha)cdot beta$.
answered Jan 27 at 17:09
YankoYanko
8,0832830
$endgroup$
Hint:
Use this fact: For all $alpha$, $-alpha = (-1)cdot alpha$.
Proof of the fact: $alpha+(-1)cdot alpha = 1cdot alpha + (-1)cdotalpha = (1-1)cdot alpha = 0cdot alpha = 0$.
The last equality holds because $0cdot alpha = (0+0)cdot alpha = 0cdotalpha + 0 cdot alpha$ and so by removing $0cdot alpha$ from both sides we get $0cdotalpha = 0$.
From this fact you can deduce all of the equalities. I'll do one of them and leave the rest for you
Example: $alpha cdot (-beta) = (-alpha)beta$
Proof: $alphacdot (-beta) = alphacdot ((-1)cdot beta) = (alpha cdot (-1))cdot beta = ((-1)cdot alpha)cdot beta = (-alpha)cdot beta$.
answered Jan 27 at 17:09
YankoYanko
8,0832830
answered Jan 27 at 17:09
YankoYanko
8,0832830
answered Jan 27 at 17:09
YankoYanko
8,0832830
answered Jan 27 at 17:09
YankoYanko
8,0832830
8,0832830
add a comment |
add a comment |
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$begingroup$
Possibly factoring out your "$-1$" through commutativity and associativity?
$endgroup$
– Hyperion
Jan 27 at 17:08
$begingroup$
You'll increase your chances of getting an answer if you include the axioms. As for a place to start, note the following: for any $x$ in the given field, $x=-(alphabeta) iff x+alphabeta=0$, this is by definition of additive inverse.
$endgroup$
– Git Gud
Jan 27 at 17:09