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10:06
1
Q: Inductive definition of a set

GomathiI'm a beginner in set theory and I have doubt regarding mathematical induction. I came across the following examples. Example 1: Find the set given by the following definition: 1) $ 3 \in P $ 2) For $x,y \in P, x + y \in P $ 3) Only those elements obtained from steps (1) and (2) are in $P$ ...

Which part do you not understand?
@DanielPietrobon How did they found that the set has multiples of 3 in example 1 and how to write the definitions in Example 2, Sir?
Well, what is 3+3?
@DanielPietrobon But, they didn't mention that y is also 3, Sir.
x and y are elements in P. 3 is an element in P by (1).
10:06
@DanielPietrobon So, initially as they have mentioned 3 belongs to P, we take both x and y as 3. Am I right, Sir?
Yes, exactly. :)
So now you know that 3 and 3+3=6 are in P.
What other numbers does P contain?
@DanielPietrobon now it has 3 and 6 sir
and what else?
Remember, for any elements x and y in P, x+y is also in P.
so, 3, 6, 9 ,12, etc
Very good! Is it clear that these are exactly the multiples of three?
10:16
yes sir. Thank you.
Great, now go back and see if Question 2 is clearer to you. :)
Good day Gomathi.
Oh, and by the way, I am no sir. ;)
@DanielPietrobon So, first time we take 3, then 3+3=6, then 3+6=9, then 9+6=15, etc. Am I correct?
Yes, but is enough to notice that 3 + 3 + 3 + ... + 3 (k times) is 3*k, a multiple of 3.
In the second example, it is given as both 2 and 3 because multiples of 2 or multiples of 3 alone doesn't make the set. So, initially they have given both. Am I correct?
Yes. That is a nice observation.
If we only had 2 we could only make 2,4,6,8,10,... etc.
These are all the even numbers.
How are we going to make all the odd numbers?
10:23
by giving multiples of 3
Does that work? What are the multiples of 3? Write the first few down.
3,6,9,...
So you are missing 5 and 7...
once 3 comes in there, we can form 5 using 3+2,...so we can get multiples of 5 also...am I correct?
Yes. Very good.
10:25
after 5, 5+2 for 7
By gosh, I think you've got it!
Yes.Also, why is it always x+y,...? Can't it be x*y or any other relations?
Because those are the rules of the game.
so, is it x+y always?
You might like to think of what would happen if you replaced they condition
x + y is in P
with x*y is in P
10:27
we will be missing certain elements
Indeed.
If we knew that 3 was in P.
and x*y was in P.
What elements would we get?
3,9,27,...
Very good. I think you understand the idea now. :)
we miss 6...
yes...thank you so much
to make it clear.
the first step is the initial set of elements
only based on the second step, we add the remaining elements to the initial step.
am i right?
Yes sir. ;)
10:31
Am a girl. :)
Can you kindly gather these together and post it as answer, so that I will accept it.
10:48
@DanielPietrobon ok brother. Thank you so much for your help.
good night

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