## mathmath333 one year ago Set Theory

1. mathmath333

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 $$\large \mathbb{Z}$$ ?

2. amilapsn

for f to be a function, what properties should f have?

3. mathmath333

1 to 1 image

4. amilapsn

$\large f=\{(ab,a+b)\ ,a+b)\ :\ a,\ b \in \mathbb{Z}\}$Check if this is correct...

5. mathmath333

correct

6. amilapsn

I think there's typo in (ab,a+b)

7. mathmath333

8. amilapsn

ok ...

9. anonymous

yes

10. amilapsn

0 maps to what?

11. mathmath333

$$\large {\text{Let} \quad f \ \text{be the subset of}\quad \mathbb{Z}\times \mathbb{Z} \\ \text{defined by} \quad f=\{(ab,a+b)\ :\ a,\ b \in \mathbb{Z}\} \\ .\text{Is}\quad f\ \text{ a function from} \ \ \mathbb{Z} \ \text{to}\ \mathbb{Z}}$$

12. mathmath333

typo corrected

13. amilapsn

so, 0 maps to what in co domain?

14. anonymous

1

15. mathmath333

i didnt understand ur question

16. amilapsn

|dw:1433162361384:dw|

17. amilapsn

ok?

18. anonymous

ab, a+b

19. amilapsn

and many other values......

20. anonymous

a neq b where Z^+

21. mathmath333

m still confuzed

22. anonymous

create a condition when it is not a one-to-one function

23. amilapsn

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?

24. mathmath333

no it has too many values to be unique.

25. anonymous

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

26. amilapsn

You got it @mathmath333

27. anonymous

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

28. amilapsn

@hwyl I just show part of domain and codomain .. Not all elements...

29. amilapsn

In fact they are infinite sets...

30. anonymous

so when is it not a function?

31. mathmath333

so it is not a function ?

32. mathmath333

@amilapsn

33. amilapsn

you got... Do you want me to explain it little deeper?

34. mathmath333

no

35. amilapsn

okay