Is there a bound for $|a+b|$ of the form $f(a,b) leq |a+b|$
0
$begingroup$
$a,b$ are integers and the $f(a,b)$ is a function of $a$ and $b$.
I know that $|a+b| leq |a|+|b|$. But what about $(? leq |a+b|)$
elementary-number-theory inequality absolute-value
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
$endgroup$
add a comment |
0
$begingroup$
$a,b$ are integers and the $f(a,b)$ is a function of $a$ and $b$.
I know that $|a+b| leq |a|+|b|$. But what about $(? leq |a+b|)$
elementary-number-theory inequality absolute-value
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
$endgroup$
$begingroup$
$|a+b| le |a|+|b| implies |x+y-y| le |x+y|+|-y|$.
$endgroup$
– Math Lover
Dec 18 '18 at 17:54
$begingroup$
That would be $0leq |a+b|$, which is also the best possible since you can have $b=-a$.
$endgroup$
– Yong Hao Ng
Dec 19 '18 at 2:27
add a comment |
0
0
0
$begingroup$
$a,b$ are integers and the $f(a,b)$ is a function of $a$ and $b$.
I know that $|a+b| leq |a|+|b|$. But what about $(? leq |a+b|)$
elementary-number-theory inequality absolute-value
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
$endgroup$
$a,b$ are integers and the $f(a,b)$ is a function of $a$ and $b$.
I know that $|a+b| leq |a|+|b|$. But what about $(? leq |a+b|)$
elementary-number-theory inequality absolute-value
elementary-number-theory inequality absolute-value
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
asked Dec 18 '18 at 17:25
PintecoPinteco
761313
761313
$begingroup$
$|a+b| le |a|+|b| implies |x+y-y| le |x+y|+|-y|$.
$endgroup$
– Math Lover
Dec 18 '18 at 17:54
$begingroup$
That would be $0leq |a+b|$, which is also the best possible since you can have $b=-a$.
$endgroup$
– Yong Hao Ng
Dec 19 '18 at 2:27
add a comment |
$begingroup$
$|a+b| le |a|+|b| implies |x+y-y| le |x+y|+|-y|$.
$endgroup$
– Math Lover
Dec 18 '18 at 17:54
$begingroup$
That would be $0leq |a+b|$, which is also the best possible since you can have $b=-a$.
$endgroup$
– Yong Hao Ng
Dec 19 '18 at 2:27
$begingroup$
$|a+b| le |a|+|b| implies |x+y-y| le |x+y|+|-y|$.
$endgroup$
– Math Lover
Dec 18 '18 at 17:54
$begingroup$
$|a+b| le |a|+|b| implies |x+y-y| le |x+y|+|-y|$.
$endgroup$
– Math Lover
Dec 18 '18 at 17:54
$begingroup$
That would be $0leq |a+b|$, which is also the best possible since you can have $b=-a$.
$endgroup$
– Yong Hao Ng
Dec 19 '18 at 2:27
$begingroup$
That would be $0leq |a+b|$, which is also the best possible since you can have $b=-a$.
$endgroup$
– Yong Hao Ng
Dec 19 '18 at 2:27
add a comment |
1 Answer
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$begingroup$
You are thinking of the reverse triangle inequality: $$||a|-|b|| leq |a-b|.$$
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
$endgroup$
$begingroup$
No @amWhy. The OP stated that "they know $|a+b|le |a|+|b|$", confusing the triangle inequality (the one they provide is just an equality). The OP isn't asking about $||a|-|b||$, nor is D.B. suggesting that...
$endgroup$
– Rhys Hughes
Dec 18 '18 at 17:33
$begingroup$
Got it; sorry for my confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:34
$begingroup$
@RhysHughes The comment above, I guess, should be directed to you. I thought it was from D.B. In any case, to both of you, sorry for the confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:41
$begingroup$
Is this valid for real numbers $a,b$? Why the minus sign at $|a-b|$, could I just change to $|a+b|$?
$endgroup$
– Pinteco
Dec 18 '18 at 17:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
You are thinking of the reverse triangle inequality: $$||a|-|b|| leq |a-b|.$$
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
$endgroup$
$begingroup$
No @amWhy. The OP stated that "they know $|a+b|le |a|+|b|$", confusing the triangle inequality (the one they provide is just an equality). The OP isn't asking about $||a|-|b||$, nor is D.B. suggesting that...
$endgroup$
– Rhys Hughes
Dec 18 '18 at 17:33
$begingroup$
Got it; sorry for my confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:34
$begingroup$
@RhysHughes The comment above, I guess, should be directed to you. I thought it was from D.B. In any case, to both of you, sorry for the confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:41
$begingroup$
Is this valid for real numbers $a,b$? Why the minus sign at $|a-b|$, could I just change to $|a+b|$?
$endgroup$
– Pinteco
Dec 18 '18 at 17:45
add a comment |
1
$begingroup$
You are thinking of the reverse triangle inequality: $$||a|-|b|| leq |a-b|.$$
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
$endgroup$
$begingroup$
No @amWhy. The OP stated that "they know $|a+b|le |a|+|b|$", confusing the triangle inequality (the one they provide is just an equality). The OP isn't asking about $||a|-|b||$, nor is D.B. suggesting that...
$endgroup$
– Rhys Hughes
Dec 18 '18 at 17:33
$begingroup$
Got it; sorry for my confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:34
$begingroup$
@RhysHughes The comment above, I guess, should be directed to you. I thought it was from D.B. In any case, to both of you, sorry for the confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:41
$begingroup$
Is this valid for real numbers $a,b$? Why the minus sign at $|a-b|$, could I just change to $|a+b|$?
$endgroup$
– Pinteco
Dec 18 '18 at 17:45
add a comment |
1
1
1
$begingroup$
You are thinking of the reverse triangle inequality: $$||a|-|b|| leq |a-b|.$$
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
$endgroup$
You are thinking of the reverse triangle inequality: $$||a|-|b|| leq |a-b|.$$
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
answered Dec 18 '18 at 17:27
D.B.D.B.
1,27518
1,27518
$begingroup$
No @amWhy. The OP stated that "they know $|a+b|le |a|+|b|$", confusing the triangle inequality (the one they provide is just an equality). The OP isn't asking about $||a|-|b||$, nor is D.B. suggesting that...
$endgroup$
– Rhys Hughes
Dec 18 '18 at 17:33
$begingroup$
Got it; sorry for my confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:34
$begingroup$
@RhysHughes The comment above, I guess, should be directed to you. I thought it was from D.B. In any case, to both of you, sorry for the confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:41
$begingroup$
Is this valid for real numbers $a,b$? Why the minus sign at $|a-b|$, could I just change to $|a+b|$?
$endgroup$
– Pinteco
Dec 18 '18 at 17:45
add a comment |
$begingroup$
No @amWhy. The OP stated that "they know $|a+b|le |a|+|b|$", confusing the triangle inequality (the one they provide is just an equality). The OP isn't asking about $||a|-|b||$, nor is D.B. suggesting that...
$endgroup$
– Rhys Hughes
Dec 18 '18 at 17:33
$begingroup$
Got it; sorry for my confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:34
$begingroup$
@RhysHughes The comment above, I guess, should be directed to you. I thought it was from D.B. In any case, to both of you, sorry for the confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:41
$begingroup$
Is this valid for real numbers $a,b$? Why the minus sign at $|a-b|$, could I just change to $|a+b|$?
$endgroup$
– Pinteco
Dec 18 '18 at 17:45
$begingroup$
No @amWhy. The OP stated that "they know $|a+b|le |a|+|b|$", confusing the triangle inequality (the one they provide is just an equality). The OP isn't asking about $||a|-|b||$, nor is D.B. suggesting that...
$endgroup$
– Rhys Hughes
Dec 18 '18 at 17:33
$begingroup$
No @amWhy. The OP stated that "they know $|a+b|le |a|+|b|$", confusing the triangle inequality (the one they provide is just an equality). The OP isn't asking about $||a|-|b||$, nor is D.B. suggesting that...
$endgroup$
– Rhys Hughes
Dec 18 '18 at 17:33
$begingroup$
Got it; sorry for my confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:34
$begingroup$
Got it; sorry for my confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:34
$begingroup$
@RhysHughes The comment above, I guess, should be directed to you. I thought it was from D.B. In any case, to both of you, sorry for the confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:41
$begingroup$
@RhysHughes The comment above, I guess, should be directed to you. I thought it was from D.B. In any case, to both of you, sorry for the confusion.
$endgroup$
– Namaste
Dec 18 '18 at 17:41
$begingroup$
Is this valid for real numbers $a,b$? Why the minus sign at $|a-b|$, could I just change to $|a+b|$?
$endgroup$
– Pinteco
Dec 18 '18 at 17:45
$begingroup$
Is this valid for real numbers $a,b$? Why the minus sign at $|a-b|$, could I just change to $|a+b|$?
$endgroup$
– Pinteco
Dec 18 '18 at 17:45
add a comment |
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$begingroup$
$|a+b| le |a|+|b| implies |x+y-y| le |x+y|+|-y|$.
$endgroup$
– Math Lover
Dec 18 '18 at 17:54
$begingroup$
That would be $0leq |a+b|$, which is also the best possible since you can have $b=-a$.
$endgroup$
– Yong Hao Ng
Dec 19 '18 at 2:27