Question

Which of these sets have the inductive property:
1) {20,21,22,23,...}
2){1,2,4,5,6,7,...}
3) {\( x \in N : x^2 \le 1000 \)}

Answer 1

1). It has a general pattern that allows us to predict the next number in the set.

Answer 2

what abt #3?

Answer 3

Wait....I'm sorry. I think I got this confused with induction....sorry I don't know :(.

Answer 4

nooo. A set S of natural numbers is inductive iff it has the property that whenever \( n \in S \) then \( n+1 \in S \)

Answer 5

@SmoothMath can u help me?

Answer 6

I know that #1 is but there seems to be another set by the way the question was phrased/ So u think the #3 is an option too?

Answer 7

Swissgirl, let's look at #3 and ask some questions.
Is 31 a members of that set?

Answer 8

nope

Answer 9

orly? Tell me why.

Answer 10

since there is no natural number,x that when u square it will give you 31

Answer 11

You are misreading that set.

Answer 12

ohhhhh yaaaaaa ok i did lol

Answer 13

{x∈N:x^2≤1000}
means
"x in N such that x^2 is less than or equal to 1000"
That means that x is in the set if it satisfies those conditions. It has to be in N, that is, a natural number. And it has to square to be less than or equal to 1000.

Answer 14

yaaaaa ok so then its inductive

Answer 15

D':

Answer 16

but 31 isnt part of the set

Answer 17

since 31^2=961

Answer 18

31 isn't in the set, eh?

Answer 19

again i misread even what u wrote lol

Answer 20

so then its not inductive since it doesnt go on forever

Answer 21

Swissgirl, there are two things that we need to understand here before we can answer this question, and I don't think you understand either of them.
First we have to understand what it means for a a set to be inductive.
And we need to understand what the set is.

Answer 22

Take it slow and understand both of those things completely before you try to jump to an answer.

Answer 23

Your understanding of inductive is actually pretty good. "It goes on forever" pretty much works. If there is a last element, then it is not inductive.

Answer 24

ya since if n is the last number then there wont be an n+1

Answer 25

Good good.

Answer 26

so then its not inductive

Answer 27

Explain to me your understanding of the set.

Answer 28

lets call this set S which is composed of natural numbers. Whenever \( n \in S \) then \(n+1 \in S \)

Answer 29

Oh, I meant your understanding of what set 3 is.

Answer 30

ohhhhh lol

Answer 31

ok so the set S includes all natural numbers that when squared is less or equal to 1000

Answer 32

is less than *

Answer 33

so let n be an element of this set S. Let n=31 so 31^2=961
now n+1=32 and 32^2=1024> since n+1 isnt an element of this set so this set isnt inductive

Answer 34

Very good =)

Answer 35

YAAYYYYYYYYYYY. Thanks

Answer 36

So for any set, if n is in the set, but n+1 is not, then that set is not inductive.

Answer 37

yuppp gotchhaaa

Answer 38

Which means that if some number is in the set, then every number after it has to be too. So as you said, it has to go on forever, at least from the starting point that you establish.

Answer 39

Thanks :)

Answer 40

Mhmm =)
Question. Is it the notation that they used for that third set that confused you? Can I help you out with how to read that kind of notation?

Answer 41

umm well kinda. I do have difficulties reading it plus i was chatting at the same time to be honest

Answer 42

Okay, I understand. Just take your time and try to wrap your head around what the notation means.
The : in the notation means "such that" and what comes after it should be thought of as a criteria for what can be included in the set.

Answer 43

okkkkkkkkkk gotcha thanks

Answer 44

Yup =)