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Ex: take $a = 2$ and $b = 3$

if the expression is $a+b$

$$a+b = 5\\b+a = 5$$ too. (which i do not need)

but what I need is $b+a$ should not be equal to $5 $

expression (a,b) != (is not equal to) expression(b,a) and Σ expression (x,y) (x= {0,1,2,3,---} - {a} and y = {0,1,2,3,---} - {b} )

so if you interchange the places of b and a then the value should be different to the value it is before.

kingW3
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vamsi
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    What you are talking about is called commutativity. Two elements $a$ and $b$ are called commutative is $a \circ b = b \circ a$ for $\circ$ being an operator. A set $X$ is called commutative if $\forall \ a, b \in X$, $a \circ b = b \circ a$. If the set isn't commutative then it is non-commutative. – Chinmay Nirkhe Jan 02 '15 at 19:09
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    What you are looking for is a noncommutative set. Try $a \circ b = a - b$ or for a more complicated example try $a \circ b = a^b$. In this case $a \circ b = a^b$ and $b \circ a = b^a$ which are not necessarily equal. – Chinmay Nirkhe Jan 02 '15 at 19:10
  • Also refer to the LaTeX guide when next writing your question. – Chinmay Nirkhe Jan 02 '15 at 19:11
  • Here is the latex guide http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Asinomás Jan 02 '15 at 19:13
  • See here http://math.stackexchange.com/questions/75107/injective-map-from-mathbbr2-to-mathbbr – spin Jan 02 '15 at 19:15
  • Can you try to make it clearer what you are asking? I am not sure if what you want is a function $f$ such that $f(a,b)$ is unique for any pair $(a,b)$, or a function $f$ such that $f(a,b) = f(b,a)$ implies $a = b$. – Caleb Stanford Jan 02 '15 at 19:47

3 Answers3

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Ok, so what you want seems to be an operation $\star$ on the real numbers that such that $a\star b=b\star a$ happens only when $a=b$.

An example of such an operation is subtraction.

As another example consider the right identity operation, defined by $a\star b=a$

Asinomás
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Since you have not specified a domain and codomain or range for this function, the answer is an easy yes. Just take the function $f:\Bbb{R}^2\to\Bbb{C}$ where $f(a,b)=a+bi$.

If you are looking for a function that is an injection from $\Bbb{R}^2$ to $\Bbb{R}$, that is, where every distinct ordered pair input has a unique real-number output, the answer is still yes, because $|\Bbb{R}^2|=|\Bbb{R}|$. However, finding a specific function is a little more complicated. I suggest looking at this or this question, which deals with this.

Community
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KSmarts
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What you are looking for seems to be an injection from $R^2$ to $R$.

Dasherman
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