Intro to binary operations
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Consider the closed interval $[0,1]$ and define $*$ on $[0,1]$ via $a*b=min[a,b]$. Determine whether $*$ is a binary operation on $[0,1]$.
No formal answer yet -- just throwing ideas around and need help with my scattered thoughts.
Since it is a closed interval $[0,1]$, the minimum for $a$ and $b$ would be $0,1$ as long as $aneq b$?
Or is there something I am missing... Sorry, we started binary operations today so I am a little unsure on the concepts.
EDIT:
If its a closed interval $[0,1]$, any operation $a,b$ yields cannot be outside the interval(obviously?) Just putting together random thoughts. Again, sorry if its me just stating obvious, rudimentary things.
abstract-algebra elementary-number-theory
asked Jan 13 at 2:56
RyanRyan
1356
$endgroup$
|
show 4 more comments
$begingroup$
Consider the closed interval $[0,1]$ and define $*$ on $[0,1]$ via $a*b=min[a,b]$. Determine whether $*$ is a binary operation on $[0,1]$.
No formal answer yet -- just throwing ideas around and need help with my scattered thoughts.
Since it is a closed interval $[0,1]$, the minimum for $a$ and $b$ would be $0,1$ as long as $aneq b$?
Or is there something I am missing... Sorry, we started binary operations today so I am a little unsure on the concepts.
EDIT:
If its a closed interval $[0,1]$, any operation $a,b$ yields cannot be outside the interval(obviously?) Just putting together random thoughts. Again, sorry if its me just stating obvious, rudimentary things.
abstract-algebra elementary-number-theory
asked Jan 13 at 2:56
RyanRyan
1356
$endgroup$
$begingroup$
There are no occurrences of $a,b$ in $min [0,1].$ Is this intentional? Or do you mean a * b to be $min(a,b)$ for any $a,b in [0,1]$ ?
$endgroup$
– coffeemath
Jan 13 at 3:08
$begingroup$
Thanks for your reply, but I am not understanding what you mean. That is what the problem states directly from the book. "via $a*b=min[a,b]$. In other words, take the minimum of $a$ and $b$.
$endgroup$
– Ryan
Jan 13 at 3:12
1
$begingroup$
If it says "via a*b = $min[a,b]$ then you changed definition when you used $0,1$ instead of $a,b$ on the right of definition.
$endgroup$
– coffeemath
Jan 13 at 3:15
1
$begingroup$
Ah! I gotcha. Thanks! I totally typo'd that. Sorry!
$endgroup$
– Ryan
Jan 13 at 3:16
3
$begingroup$
It's not about closed intervals. If $S$ is any set of real numbers, then $$a*b=min(a,b)$$ defines a binary operation on $S$, because the minimum of two numbers is one of those numbers. To put it more pedantically, $$a*b=min(a,b)in{a,b}subseteq S.$$
$endgroup$
– bof
Jan 13 at 3:23
|
show 4 more comments
$begingroup$
Consider the closed interval $[0,1]$ and define $*$ on $[0,1]$ via $a*b=min[a,b]$. Determine whether $*$ is a binary operation on $[0,1]$.
No formal answer yet -- just throwing ideas around and need help with my scattered thoughts.
Since it is a closed interval $[0,1]$, the minimum for $a$ and $b$ would be $0,1$ as long as $aneq b$?
Or is there something I am missing... Sorry, we started binary operations today so I am a little unsure on the concepts.
EDIT:
If its a closed interval $[0,1]$, any operation $a,b$ yields cannot be outside the interval(obviously?) Just putting together random thoughts. Again, sorry if its me just stating obvious, rudimentary things.
abstract-algebra elementary-number-theory
asked Jan 13 at 2:56
RyanRyan
1356
$endgroup$
Consider the closed interval $[0,1]$ and define $*$ on $[0,1]$ via $a*b=min[a,b]$. Determine whether $*$ is a binary operation on $[0,1]$.
No formal answer yet -- just throwing ideas around and need help with my scattered thoughts.
Since it is a closed interval $[0,1]$, the minimum for $a$ and $b$ would be $0,1$ as long as $aneq b$?
Or is there something I am missing... Sorry, we started binary operations today so I am a little unsure on the concepts.
EDIT:
If its a closed interval $[0,1]$, any operation $a,b$ yields cannot be outside the interval(obviously?) Just putting together random thoughts. Again, sorry if its me just stating obvious, rudimentary things.
abstract-algebra elementary-number-theory
abstract-algebra elementary-number-theory
asked Jan 13 at 2:56
RyanRyan
1356
asked Jan 13 at 2:56
RyanRyan
1356
edited Jan 13 at 3:16
Ryan
asked Jan 13 at 2:56
RyanRyan
1356
asked Jan 13 at 2:56
RyanRyan
1356
asked Jan 13 at 2:56
RyanRyan
1356
1356
$begingroup$
There are no occurrences of $a,b$ in $min [0,1].$ Is this intentional? Or do you mean a * b to be $min(a,b)$ for any $a,b in [0,1]$ ?
$endgroup$
– coffeemath
Jan 13 at 3:08
$begingroup$
Thanks for your reply, but I am not understanding what you mean. That is what the problem states directly from the book. "via $a*b=min[a,b]$. In other words, take the minimum of $a$ and $b$.
$endgroup$
– Ryan
Jan 13 at 3:12
1
$begingroup$
If it says "via a*b = $min[a,b]$ then you changed definition when you used $0,1$ instead of $a,b$ on the right of definition.
$endgroup$
– coffeemath
Jan 13 at 3:15
1
$begingroup$
Ah! I gotcha. Thanks! I totally typo'd that. Sorry!
$endgroup$
– Ryan
Jan 13 at 3:16
3
$begingroup$
It's not about closed intervals. If $S$ is any set of real numbers, then $$a*b=min(a,b)$$ defines a binary operation on $S$, because the minimum of two numbers is one of those numbers. To put it more pedantically, $$a*b=min(a,b)in{a,b}subseteq S.$$
$endgroup$
– bof
Jan 13 at 3:23
|
show 4 more comments
$begingroup$
There are no occurrences of $a,b$ in $min [0,1].$ Is this intentional? Or do you mean a * b to be $min(a,b)$ for any $a,b in [0,1]$ ?
$endgroup$
– coffeemath
Jan 13 at 3:08
$begingroup$
Thanks for your reply, but I am not understanding what you mean. That is what the problem states directly from the book. "via $a*b=min[a,b]$. In other words, take the minimum of $a$ and $b$.
$endgroup$
– Ryan
Jan 13 at 3:12
1
$begingroup$
If it says "via a*b = $min[a,b]$ then you changed definition when you used $0,1$ instead of $a,b$ on the right of definition.
$endgroup$
– coffeemath
Jan 13 at 3:15
1
$begingroup$
Ah! I gotcha. Thanks! I totally typo'd that. Sorry!
$endgroup$
– Ryan
Jan 13 at 3:16
3
$begingroup$
It's not about closed intervals. If $S$ is any set of real numbers, then $$a*b=min(a,b)$$ defines a binary operation on $S$, because the minimum of two numbers is one of those numbers. To put it more pedantically, $$a*b=min(a,b)in{a,b}subseteq S.$$
$endgroup$
– bof
Jan 13 at 3:23
$begingroup$
There are no occurrences of $a,b$ in $min [0,1].$ Is this intentional? Or do you mean a * b to be $min(a,b)$ for any $a,b in [0,1]$ ?
$endgroup$
– coffeemath
Jan 13 at 3:08
$begingroup$
There are no occurrences of $a,b$ in $min [0,1].$ Is this intentional? Or do you mean a * b to be $min(a,b)$ for any $a,b in [0,1]$ ?
$endgroup$
– coffeemath
Jan 13 at 3:08
$begingroup$
Thanks for your reply, but I am not understanding what you mean. That is what the problem states directly from the book. "via $a*b=min[a,b]$. In other words, take the minimum of $a$ and $b$.
$endgroup$
– Ryan
Jan 13 at 3:12
$begingroup$
Thanks for your reply, but I am not understanding what you mean. That is what the problem states directly from the book. "via $a*b=min[a,b]$. In other words, take the minimum of $a$ and $b$.
$endgroup$
– Ryan
Jan 13 at 3:12
1
1
$begingroup$
If it says "via a*b = $min[a,b]$ then you changed definition when you used $0,1$ instead of $a,b$ on the right of definition.
$endgroup$
– coffeemath
Jan 13 at 3:15
$begingroup$
If it says "via a*b = $min[a,b]$ then you changed definition when you used $0,1$ instead of $a,b$ on the right of definition.
$endgroup$
– coffeemath
Jan 13 at 3:15
1
1
$begingroup$
Ah! I gotcha. Thanks! I totally typo'd that. Sorry!
$endgroup$
– Ryan
Jan 13 at 3:16
$begingroup$
Ah! I gotcha. Thanks! I totally typo'd that. Sorry!
$endgroup$
– Ryan
Jan 13 at 3:16
3
3
$begingroup$
It's not about closed intervals. If $S$ is any set of real numbers, then $$a*b=min(a,b)$$ defines a binary operation on $S$, because the minimum of two numbers is one of those numbers. To put it more pedantically, $$a*b=min(a,b)in{a,b}subseteq S.$$
$endgroup$
– bof
Jan 13 at 3:23
$begingroup$
It's not about closed intervals. If $S$ is any set of real numbers, then $$a*b=min(a,b)$$ defines a binary operation on $S$, because the minimum of two numbers is one of those numbers. To put it more pedantically, $$a*b=min(a,b)in{a,b}subseteq S.$$
$endgroup$
– bof
Jan 13 at 3:23
|
show 4 more comments
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$begingroup$
There are no occurrences of $a,b$ in $min [0,1].$ Is this intentional? Or do you mean a * b to be $min(a,b)$ for any $a,b in [0,1]$ ?
$endgroup$
– coffeemath
Jan 13 at 3:08
$begingroup$
Thanks for your reply, but I am not understanding what you mean. That is what the problem states directly from the book. "via $a*b=min[a,b]$. In other words, take the minimum of $a$ and $b$.
$endgroup$
– Ryan
Jan 13 at 3:12
1
$begingroup$
If it says "via a*b = $min[a,b]$ then you changed definition when you used $0,1$ instead of $a,b$ on the right of definition.
$endgroup$
– coffeemath
Jan 13 at 3:15
1
$begingroup$
Ah! I gotcha. Thanks! I totally typo'd that. Sorry!
$endgroup$
– Ryan
Jan 13 at 3:16
3
$begingroup$
It's not about closed intervals. If $S$ is any set of real numbers, then $$a*b=min(a,b)$$ defines a binary operation on $S$, because the minimum of two numbers is one of those numbers. To put it more pedantically, $$a*b=min(a,b)in{a,b}subseteq S.$$
$endgroup$
– bof
Jan 13 at 3:23