Set Theory

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Let \(\large f\) be the subset of \(\large \mathbb{Z}\times \mathbb{Z}\) defined by \(\large f=\{(ab,a+b)\ ,a+b)\ :\ a,\ b \in \mathbb{Z}\}\) .Is \(\large f\) a function from \(\large \mathbb{Z}\) to \(\large \mathbb{Z}\) ?
for f to be a function, what properties should f have?
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\[\large f=\{(ab,a+b)\ ,a+b)\ :\ a,\ b \in \mathbb{Z}\}\]Check if this is correct...
correct
I think there's typo in (ab,a+b)
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ok ...
yes
0 maps to what?
\(\large {\text{Let} \quad f \ \text{be the subset of}\quad \mathbb{Z}\times \mathbb{Z} \\ \text{defined by} \quad f=\{(ab,a+b)\ :\ a,\ b \in \mathbb{Z}\} \\ .\text{Is}\quad f\ \text{ a function from} \ \ \mathbb{Z} \ \text{to}\ \mathbb{Z}} \)
typo corrected
so, 0 maps to what in co domain?
1
i didnt understand ur question
|dw:1433162361384:dw|
ok?
ab, a+b
and many other values......
a neq b where Z^+
m still confuzed
create a condition when it is not a one-to-one function
by the definition of f, we can choose any a, b in integer set. So f will include ordered pairs: (0,1) when a=0 and b=1 (0,2) when a=0 and b=2 and so on.... So to be a function every x value should have a unique y value... So will f be a function?
no it has too many values to be unique.
you don't make any sense she clearly knows the definition of a function under what condition that (ab, a+b) is not a function when a,b are elements of Z
You got it @mathmath333
if your domain is 0, clearly either a = 0 or b = 0 and one of them has to be greater than zero [(0*1), (0+1)] domain is 0, but the range is 1 and not 1 and 2
@hwyl I just show part of domain and codomain .. Not all elements...
In fact they are infinite sets...
so when is it not a function?
so it is not a function ?
you got... Do you want me to explain it little deeper?
no
okay

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