- #1
SithsNGiggles
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Homework Statement
The objective was to think of a binary operation ##*:\mathbb{N}\times\mathbb{N}\to\mathbb{N}## that is injective. A classmate came up with the following operation, but had trouble showing it was injective:
##a*b=a^3+b^4##.
Homework Equations
The Attempt at a Solution
Here's my thought process. I denote ##a*b## and ##c*d## by ##f(a,b)## and ##f(c,d)##, respectively. The usual procedure to show a function is injective is by using the definition:
A function ##f## is injective if ##f(a,b)=f(c,d)~\Rightarrow~(a,b)=(c,d)##. Alternatively, ##f## is injective if ##(a,b)\not=(c,d)~\Rightarrow~f(a,b)\not=f(c,d)##.
I figure the second (contrapositive) statement would be easier to prove. I assume ##(a,b)\not=(c,d)##; that is, I assume ##a\not=c## OR ##b\not=d##.
Suppose ##a\not=c,b=d##.
##\begin{align*}
a&\not=c\\
a^3&\not=c^3\\
a^3+b^4&\not=c^3+b^4\\
a^3+b^4&\not=c^3+d^4\\
a*b&\not=c*d\end{align*}##
Suppose ##a=c,b\not=d##.
##\begin{align*}
b&\not=d\\
b^4&\not=d^4\\
a^3+b^4&\not=a^3+d^4\\
a^3+b^4&\not=c^3+d^4\\
a*b&\not=c*d\end{align*}##
Now, I'm under the impression that this is all I have to show in order to conclude that ##*## is injective. Am I right in thinking this way?