Mixing
Proposition: For all $m, n, p, q$ contained in Natural Numbers, $(m-n)p = mp - np$
$begingroup$
I have to prove the following proposition using the axioms for integers, and the definition of subtraction to be $m - n = m + (-n).$
Proof:
$(m-n)p = (m+(-n))p$ by definition of subtraction
$(m+(-n))p = mp + (-n)p$ by Distributivity
$mp + (-n)p = mp - np$ by definition of subtraction
My last line was marked wrong; is this not a valid step to finish the proof?
elementary-number-theory
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
$endgroup$
add a comment |
$begingroup$
I have to prove the following proposition using the axioms for integers, and the definition of subtraction to be $m - n = m + (-n).$
Proof:
$(m-n)p = (m+(-n))p$ by definition of subtraction
$(m+(-n))p = mp + (-n)p$ by Distributivity
$mp + (-n)p = mp - np$ by definition of subtraction
My last line was marked wrong; is this not a valid step to finish the proof?
elementary-number-theory
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
$endgroup$
$begingroup$
You need to prove that $(-n)p=-(np)$. You can do that by adding $np $ to each side to get $(-n)p+np=-(np)+np $ so $(-n+n)p=0$ and then you have to prove $0p=0$.
$endgroup$
– fleablood
Feb 3 at 1:34
add a comment |
$begingroup$
I have to prove the following proposition using the axioms for integers, and the definition of subtraction to be $m - n = m + (-n).$
Proof:
$(m-n)p = (m+(-n))p$ by definition of subtraction
$(m+(-n))p = mp + (-n)p$ by Distributivity
$mp + (-n)p = mp - np$ by definition of subtraction
My last line was marked wrong; is this not a valid step to finish the proof?
elementary-number-theory
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
$endgroup$
I have to prove the following proposition using the axioms for integers, and the definition of subtraction to be $m - n = m + (-n).$
Proof:
$(m-n)p = (m+(-n))p$ by definition of subtraction
$(m+(-n))p = mp + (-n)p$ by Distributivity
$mp + (-n)p = mp - np$ by definition of subtraction
My last line was marked wrong; is this not a valid step to finish the proof?
elementary-number-theory
elementary-number-theory
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
edited Feb 3 at 3:15
J. W. Tanner
4,7871420
4,7871420
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
asked Feb 3 at 1:23
beepbeepboop123123beepbeepboop123123
606
606
$begingroup$
You need to prove that $(-n)p=-(np)$. You can do that by adding $np $ to each side to get $(-n)p+np=-(np)+np $ so $(-n+n)p=0$ and then you have to prove $0p=0$.
$endgroup$
– fleablood
Feb 3 at 1:34
add a comment |
$begingroup$
You need to prove that $(-n)p=-(np)$. You can do that by adding $np $ to each side to get $(-n)p+np=-(np)+np $ so $(-n+n)p=0$ and then you have to prove $0p=0$.
$endgroup$
– fleablood
Feb 3 at 1:34
$begingroup$
You need to prove that $(-n)p=-(np)$. You can do that by adding $np $ to each side to get $(-n)p+np=-(np)+np $ so $(-n+n)p=0$ and then you have to prove $0p=0$.
$endgroup$
– fleablood
Feb 3 at 1:34
$begingroup$
You need to prove that $(-n)p=-(np)$. You can do that by adding $np $ to each side to get $(-n)p+np=-(np)+np $ so $(-n+n)p=0$ and then you have to prove $0p=0$.
$endgroup$
– fleablood
Feb 3 at 1:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
What you did on the last line was to multiply $(-n)$ with $m$, which is not using the definition of subtraction
I am guessing you have some other principle that states that $(-x)y=(-xy)$, which you can use to justify that:
$mp+(-n)p=mp+(-np)$
And then you use the definition of subtraction to say that:
$mp+(-np)=mp-np$
so that you end up with:
$mp+(-n)p=mp-np$
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
What you did on the last line was to multiply $(-n)$ with $m$, which is not using the definition of subtraction
I am guessing you have some other principle that states that $(-x)y=(-xy)$, which you can use to justify that:
$mp+(-n)p=mp+(-np)$
And then you use the definition of subtraction to say that:
$mp+(-np)=mp-np$
so that you end up with:
$mp+(-n)p=mp-np$
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
$endgroup$
add a comment |
$begingroup$
What you did on the last line was to multiply $(-n)$ with $m$, which is not using the definition of subtraction
I am guessing you have some other principle that states that $(-x)y=(-xy)$, which you can use to justify that:
$mp+(-n)p=mp+(-np)$
And then you use the definition of subtraction to say that:
$mp+(-np)=mp-np$
so that you end up with:
$mp+(-n)p=mp-np$
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
$endgroup$
add a comment |
$begingroup$
What you did on the last line was to multiply $(-n)$ with $m$, which is not using the definition of subtraction
I am guessing you have some other principle that states that $(-x)y=(-xy)$, which you can use to justify that:
$mp+(-n)p=mp+(-np)$
And then you use the definition of subtraction to say that:
$mp+(-np)=mp-np$
so that you end up with:
$mp+(-n)p=mp-np$
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
$endgroup$
What you did on the last line was to multiply $(-n)$ with $m$, which is not using the definition of subtraction
I am guessing you have some other principle that states that $(-x)y=(-xy)$, which you can use to justify that:
$mp+(-n)p=mp+(-np)$
And then you use the definition of subtraction to say that:
$mp+(-np)=mp-np$
so that you end up with:
$mp+(-n)p=mp-np$
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
answered Feb 3 at 1:26
Bram28Bram28
64.5k44793
64.5k44793
add a comment |
add a comment |
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$begingroup$
You need to prove that $(-n)p=-(np)$. You can do that by adding $np $ to each side to get $(-n)p+np=-(np)+np $ so $(-n+n)p=0$ and then you have to prove $0p=0$.
$endgroup$
– fleablood
Feb 3 at 1:34