If $a<b$, $c<d$ then $a+c<b+d$?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
This is how I managed to prove it:
I know $a lt b$ and $c lt d$ thus, $b-a$ and $d-c$ are real positive numbers. Then, $b-a + d-c gt 0$ and because of this $a+c lt b+d$
Did I prove it right?
calculus proof-verification inequality real-numbers
asked Aug 20 '17 at 22:05
TheNicouU
169111
 |Â
show 3 more comments
up vote
2
down vote
favorite
This is how I managed to prove it:
I know $a lt b$ and $c lt d$ thus, $b-a$ and $d-c$ are real positive numbers. Then, $b-a + d-c gt 0$ and because of this $a+c lt b+d$
Did I prove it right?
calculus proof-verification inequality real-numbers
asked Aug 20 '17 at 22:05
TheNicouU
169111
It seems correct to me!
– Jaideep Khare
Aug 20 '17 at 22:05
You are using that $x gt y iff x - y gt 0,$, and $x gt 0, y gt 0 implies x+y gt 0,$. Is that something you proved before, or are otherwise allowed to use?
– dxiv
Aug 20 '17 at 22:07
I proved that before. But im going to include it in the proof. Thank you.
– TheNicouU
Aug 20 '17 at 22:09
@TheNicouU If you happen to have also proved before that $x gt y implies x+z gt y+z,$, then you can just use that twice: $a+c lt b+c lt b+d$.
– dxiv
Aug 20 '17 at 22:12
I'd just try something along the lines of if a<b than b=xa for some value x>1, if c<d then d=yc for some value y>1 so we can rewrite the other inequality as a+c<xa+yc and with x,y >1 ...
– user451844
Aug 20 '17 at 22:14
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is how I managed to prove it:
I know $a lt b$ and $c lt d$ thus, $b-a$ and $d-c$ are real positive numbers. Then, $b-a + d-c gt 0$ and because of this $a+c lt b+d$
Did I prove it right?
calculus proof-verification inequality real-numbers
asked Aug 20 '17 at 22:05
TheNicouU
169111
This is how I managed to prove it:
I know $a lt b$ and $c lt d$ thus, $b-a$ and $d-c$ are real positive numbers. Then, $b-a + d-c gt 0$ and because of this $a+c lt b+d$
Did I prove it right?
calculus proof-verification inequality real-numbers
asked Aug 20 '17 at 22:05
TheNicouU
169111
edited Aug 26 at 20:20
asked Aug 20 '17 at 22:05
TheNicouU
169111
asked Aug 20 '17 at 22:05
TheNicouU
169111
asked Aug 20 '17 at 22:05
TheNicouU
169111
169111
It seems correct to me!
– Jaideep Khare
Aug 20 '17 at 22:05
You are using that $x gt y iff x - y gt 0,$, and $x gt 0, y gt 0 implies x+y gt 0,$. Is that something you proved before, or are otherwise allowed to use?
– dxiv
Aug 20 '17 at 22:07
I proved that before. But im going to include it in the proof. Thank you.
– TheNicouU
Aug 20 '17 at 22:09
@TheNicouU If you happen to have also proved before that $x gt y implies x+z gt y+z,$, then you can just use that twice: $a+c lt b+c lt b+d$.
– dxiv
Aug 20 '17 at 22:12
I'd just try something along the lines of if a<b than b=xa for some value x>1, if c<d then d=yc for some value y>1 so we can rewrite the other inequality as a+c<xa+yc and with x,y >1 ...
– user451844
Aug 20 '17 at 22:14
 |Â
show 3 more comments
It seems correct to me!
– Jaideep Khare
Aug 20 '17 at 22:05
You are using that $x gt y iff x - y gt 0,$, and $x gt 0, y gt 0 implies x+y gt 0,$. Is that something you proved before, or are otherwise allowed to use?
– dxiv
Aug 20 '17 at 22:07
I proved that before. But im going to include it in the proof. Thank you.
– TheNicouU
Aug 20 '17 at 22:09
@TheNicouU If you happen to have also proved before that $x gt y implies x+z gt y+z,$, then you can just use that twice: $a+c lt b+c lt b+d$.
– dxiv
Aug 20 '17 at 22:12
I'd just try something along the lines of if a<b than b=xa for some value x>1, if c<d then d=yc for some value y>1 so we can rewrite the other inequality as a+c<xa+yc and with x,y >1 ...
– user451844
Aug 20 '17 at 22:14
It seems correct to me!
– Jaideep Khare
Aug 20 '17 at 22:05
It seems correct to me!
– Jaideep Khare
Aug 20 '17 at 22:05
You are using that $x gt y iff x - y gt 0,$, and $x gt 0, y gt 0 implies x+y gt 0,$. Is that something you proved before, or are otherwise allowed to use?
– dxiv
Aug 20 '17 at 22:07
You are using that $x gt y iff x - y gt 0,$, and $x gt 0, y gt 0 implies x+y gt 0,$. Is that something you proved before, or are otherwise allowed to use?
– dxiv
Aug 20 '17 at 22:07
I proved that before. But im going to include it in the proof. Thank you.
– TheNicouU
Aug 20 '17 at 22:09
I proved that before. But im going to include it in the proof. Thank you.
– TheNicouU
Aug 20 '17 at 22:09
@TheNicouU If you happen to have also proved before that $x gt y implies x+z gt y+z,$, then you can just use that twice: $a+c lt b+c lt b+d$.
– dxiv
Aug 20 '17 at 22:12
@TheNicouU If you happen to have also proved before that $x gt y implies x+z gt y+z,$, then you can just use that twice: $a+c lt b+c lt b+d$.
– dxiv
Aug 20 '17 at 22:12
I'd just try something along the lines of if a<b than b=xa for some value x>1, if c<d then d=yc for some value y>1 so we can rewrite the other inequality as a+c<xa+yc and with x,y >1 ...
– user451844
Aug 20 '17 at 22:14
I'd just try something along the lines of if a<b than b=xa for some value x>1, if c<d then d=yc for some value y>1 so we can rewrite the other inequality as a+c<xa+yc and with x,y >1 ...
– user451844
Aug 20 '17 at 22:14
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
your way is correct,another one is
add both side $c$ to the $a<b$ so that $$a+c<b+c\ $$
then add $b$ to the $c<d$ so that $$c+b<d+b$$
finally we can conclude that $$a+c<b+d$$
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
add a comment |Â
up vote
1
down vote
Yes. Formalizing, you might write something like:
$$(a<b)land(c<d)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(0<b-a)land(0<d-c)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(d-c<b-a+(d-c))land(0<d-c)$$
$implies$ (transitivity)
$$0<b-a+(d-c)$$
$implies$ (using $x<yiff x+z<y+z$ and associativity of addition)
$$a+c<b+d$$
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value."
– Thomas Andrews
Aug 20 '17 at 22:13
@ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question.
– Shuri2060
Aug 20 '17 at 22:25
1
@ThomasAndrews To me, '$b−a$ is a positive value' $equiv$ '$b-a>0$'. ie. $a<biff 0<b-a$
– Shuri2060
Aug 20 '17 at 22:34
add a comment |Â
up vote
-1
down vote
I would like to give a sweet and simple proof:
$a$ < $b$ ---------------------(1)
$c$ < $d$ ---------------------(2)
now simply add the two equations:
$a+c$ < $b+d$
hope it helps you!
answered Jul 31 at 17:18
Lol Olo
1
This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof.
– user496634
Aug 26 at 22:03
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
your way is correct,another one is
add both side $c$ to the $a<b$ so that $$a+c<b+c\ $$
then add $b$ to the $c<d$ so that $$c+b<d+b$$
finally we can conclude that $$a+c<b+d$$
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
add a comment |Â
up vote
4
down vote
accepted
your way is correct,another one is
add both side $c$ to the $a<b$ so that $$a+c<b+c\ $$
then add $b$ to the $c<d$ so that $$c+b<d+b$$
finally we can conclude that $$a+c<b+d$$
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
your way is correct,another one is
add both side $c$ to the $a<b$ so that $$a+c<b+c\ $$
then add $b$ to the $c<d$ so that $$c+b<d+b$$
finally we can conclude that $$a+c<b+d$$
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
your way is correct,another one is
add both side $c$ to the $a<b$ so that $$a+c<b+c\ $$
then add $b$ to the $c<d$ so that $$c+b<d+b$$
finally we can conclude that $$a+c<b+d$$
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
edited Aug 20 '17 at 22:17
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
answered Aug 20 '17 at 22:12
haqnatural
20.5k72457
20.5k72457
add a comment |Â
add a comment |Â
up vote
1
down vote
Yes. Formalizing, you might write something like:
$$(a<b)land(c<d)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(0<b-a)land(0<d-c)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(d-c<b-a+(d-c))land(0<d-c)$$
$implies$ (transitivity)
$$0<b-a+(d-c)$$
$implies$ (using $x<yiff x+z<y+z$ and associativity of addition)
$$a+c<b+d$$
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value."
– Thomas Andrews
Aug 20 '17 at 22:13
@ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question.
– Shuri2060
Aug 20 '17 at 22:25
1
@ThomasAndrews To me, '$b−a$ is a positive value' $equiv$ '$b-a>0$'. ie. $a<biff 0<b-a$
– Shuri2060
Aug 20 '17 at 22:34
add a comment |Â
up vote
1
down vote
Yes. Formalizing, you might write something like:
$$(a<b)land(c<d)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(0<b-a)land(0<d-c)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(d-c<b-a+(d-c))land(0<d-c)$$
$implies$ (transitivity)
$$0<b-a+(d-c)$$
$implies$ (using $x<yiff x+z<y+z$ and associativity of addition)
$$a+c<b+d$$
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value."
– Thomas Andrews
Aug 20 '17 at 22:13
@ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question.
– Shuri2060
Aug 20 '17 at 22:25
1
@ThomasAndrews To me, '$b−a$ is a positive value' $equiv$ '$b-a>0$'. ie. $a<biff 0<b-a$
– Shuri2060
Aug 20 '17 at 22:34
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes. Formalizing, you might write something like:
$$(a<b)land(c<d)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(0<b-a)land(0<d-c)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(d-c<b-a+(d-c))land(0<d-c)$$
$implies$ (transitivity)
$$0<b-a+(d-c)$$
$implies$ (using $x<yiff x+z<y+z$ and associativity of addition)
$$a+c<b+d$$
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
Yes. Formalizing, you might write something like:
$$(a<b)land(c<d)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(0<b-a)land(0<d-c)$$
$implies$ (using $x<yiff x+z<y+z$)
$$(d-c<b-a+(d-c))land(0<d-c)$$
$implies$ (transitivity)
$$0<b-a+(d-c)$$
$implies$ (using $x<yiff x+z<y+z$ and associativity of addition)
$$a+c<b+d$$
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
edited Aug 20 '17 at 22:19
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
answered Aug 20 '17 at 22:09
Shuri2060
3,479725
3,479725
But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value."
– Thomas Andrews
Aug 20 '17 at 22:13
@ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question.
– Shuri2060
Aug 20 '17 at 22:25
1
@ThomasAndrews To me, '$b−a$ is a positive value' $equiv$ '$b-a>0$'. ie. $a<biff 0<b-a$
– Shuri2060
Aug 20 '17 at 22:34
add a comment |Â
But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value."
– Thomas Andrews
Aug 20 '17 at 22:13
@ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question.
– Shuri2060
Aug 20 '17 at 22:25
1
@ThomasAndrews To me, '$b−a$ is a positive value' $equiv$ '$b-a>0$'. ie. $a<biff 0<b-a$
– Shuri2060
Aug 20 '17 at 22:34
But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value."
– Thomas Andrews
Aug 20 '17 at 22:13
But that is not the axiom system OP is using. The OPs proof is assuming the existence of a set of "positive" values, with certain properties, and defines $a<b$ as "$b-a$ is a positive value."
– Thomas Andrews
Aug 20 '17 at 22:13
@ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question.
– Shuri2060
Aug 20 '17 at 22:25
@ThomasAndrews I don't quite understand. How do you know this isn't the axiom system the OP is using? It's not stated in the question.
– Shuri2060
Aug 20 '17 at 22:25
1
1
@ThomasAndrews To me, '$b−a$ is a positive value' $equiv$ '$b-a>0$'. ie. $a<biff 0<b-a$
– Shuri2060
Aug 20 '17 at 22:34
@ThomasAndrews To me, '$b−a$ is a positive value' $equiv$ '$b-a>0$'. ie. $a<biff 0<b-a$
– Shuri2060
Aug 20 '17 at 22:34
add a comment |Â
up vote
-1
down vote
I would like to give a sweet and simple proof:
$a$ < $b$ ---------------------(1)
$c$ < $d$ ---------------------(2)
now simply add the two equations:
$a+c$ < $b+d$
hope it helps you!
answered Jul 31 at 17:18
Lol Olo
1
This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof.
– user496634
Aug 26 at 22:03
add a comment |Â
up vote
-1
down vote
I would like to give a sweet and simple proof:
$a$ < $b$ ---------------------(1)
$c$ < $d$ ---------------------(2)
now simply add the two equations:
$a+c$ < $b+d$
hope it helps you!
answered Jul 31 at 17:18
Lol Olo
1
This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof.
– user496634
Aug 26 at 22:03
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
I would like to give a sweet and simple proof:
$a$ < $b$ ---------------------(1)
$c$ < $d$ ---------------------(2)
now simply add the two equations:
$a+c$ < $b+d$
hope it helps you!
answered Jul 31 at 17:18
Lol Olo
1
I would like to give a sweet and simple proof:
$a$ < $b$ ---------------------(1)
$c$ < $d$ ---------------------(2)
now simply add the two equations:
$a+c$ < $b+d$
hope it helps you!
answered Jul 31 at 17:18
Lol Olo
1
answered Jul 31 at 17:18
Lol Olo
1
answered Jul 31 at 17:18
Lol Olo
1
answered Jul 31 at 17:18
Lol Olo
1
1
This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof.
– user496634
Aug 26 at 22:03
add a comment |Â
This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof.
– user496634
Aug 26 at 22:03
This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof.
– user496634
Aug 26 at 22:03
This is maybe the intuition behind the OP's statement, but it is nowhere near a proper proof.
– user496634
Aug 26 at 22:03
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2400679%2fif-ab-cd-then-acbd%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
It seems correct to me!
– Jaideep Khare
Aug 20 '17 at 22:05
You are using that $x gt y iff x - y gt 0,$, and $x gt 0, y gt 0 implies x+y gt 0,$. Is that something you proved before, or are otherwise allowed to use?
– dxiv
Aug 20 '17 at 22:07
I proved that before. But im going to include it in the proof. Thank you.
– TheNicouU
Aug 20 '17 at 22:09
@TheNicouU If you happen to have also proved before that $x gt y implies x+z gt y+z,$, then you can just use that twice: $a+c lt b+c lt b+d$.
– dxiv
Aug 20 '17 at 22:12
I'd just try something along the lines of if a<b than b=xa for some value x>1, if c<d then d=yc for some value y>1 so we can rewrite the other inequality as a+c<xa+yc and with x,y >1 ...
– user451844
Aug 20 '17 at 22:14