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I am just curious to know if addition of two numbers an injective function?

Lets say $\operatorname{Sum}(a,b) = a + b$

Now is the $\operatorname{Sum}$ function an injective functions?

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Since this is homework, what have you tried? – Amit Kumar Gupta Feb 14 '13 at 6:55
Is there a difference between ?, ?? and ??? in your question? – Did Feb 14 '13 at 6:55
I really don't know how to proceed with this... The questions is much more complicated that this... so... – user62263 Feb 14 '13 at 6:59
For any fixed $b$, the function $g(x)=f(x,b)$ is injective, similarly for any fixed $a$, the function $h(y)=f(a,y)$ is injective, but the function $f(x,y)$ of two variables is not injective. – André Nicolas Feb 14 '13 at 8:16

No, it's not, because $2+2=1+3$.

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I am going to be contratrian and argue that it is injective. Consider the "curried" addition function $+ : \mathbb N \to (\mathbb N\to \mathbb N)$. For example, $+(7)$ is the function which maps each number $n$ to $n+7$, and $+(0)$ is the identity function. Then $+$ is injective because there are no distinct $a, b$ for which $+(a)$ and $+(b)$ are the same function: for every $a\ne b$, there exists a $c$ such that $+(a)(c)\ne +(b)(c)$; in fact, $c=0$ will do for any choice of $a$ and $b$.

I cheerfully invite downvotes, and encourage the original poster to ignore this answer.

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Rather than be contrarian, why not be thorough and mention all three things the question could have meant? (e.g. the function you mention is injective, the function $\mathbb{N} \times \mathbb{N} \to \mathbb{N}$ is not, and each of the various functions "+a" are injective) Or at least phrase it as "when the question is interpreted this way, then that is true"? – Hurkyl Feb 14 '13 at 6:45
The OP defines a function $\mathrm{Sum}$ and asks explicitly about that function, which is clearly $\mathbb{N}^2 \to \mathbb{N}$. – Amit Kumar Gupta Feb 14 '13 at 6:53
actually MJD, you are correct... And I guess that i have partly understood your approach – user62263 Feb 14 '13 at 7:02
@hurkyl: because Emanuele had already given a perfectly clear explanation of those points, and I felt that engaging them myself would only detract from that perfect clarity. – MJD Feb 14 '13 at 13:32

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