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mathmath333
 one year ago
Set Theory
mathmath333
 one year ago
Set Theory

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1Let \(\large f\) 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 \(\large \mathbb{Z}\) ?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2for f to be a function, what properties should f have?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2\[\large f=\{(ab,a+b)\ ,a+b)\ :\ a,\ b \in \mathbb{Z}\}\]Check if this is correct...

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2I think there's typo in (ab,a+b)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large {\text{Let} \quad f \ \text{be the subset of}\quad \mathbb{Z}\times \mathbb{Z} \\ \text{defined by} \quad f=\{(ab,a+b)\ :\ a,\ b \in \mathbb{Z}\} \\ .\text{Is}\quad f\ \text{ a function from} \ \ \mathbb{Z} \ \text{to}\ \mathbb{Z}} \)

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2so, 0 maps to what in co domain?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i didnt understand ur question

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2dw:1433162361384:dw

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2and many other values......

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0create a condition when it is not a onetoone function

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2by 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?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1no it has too many values to be unique.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you 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

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2You got it @mathmath333

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if 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

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2@hwyl I just show part of domain and codomain .. Not all elements...

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2In fact they are infinite sets...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so when is it not a function?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1so it is not a function ?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.2you got... Do you want me to explain it little deeper?
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