jeetu_vishnoi wrote:A set of numbers has the property that for any number t in the set, t+2 is in the set. If -1 is in the set, which one of the following must also be in the set?
i. -3
ii. 1
iii. 5

(A) i only
(B) ii only
(C) i and ii only
(D) ii and iii only
(E) i, ii and iii

Hi there!

From the fact that -1 is in the set, you know that -1+2 = 1 is also in the set; but then 1 IS in the set, therefore 1+2 = 3 also! Now we know that 3 IS in the set, therefore 3+2 = 5 belongs to the set, too!

Till now I´ve proved that ii. and iii. are correct; what about i. ? NO!

Reason: it is POSSIBLE that -1 is the LOWEST element in the set... if so, from the property given, we know that all terms are GREATER than -1 necessarily, because t -> t+2 does not "go backwards"... that means that we cannot say whether -3 belongs to the set or not, so the answer is D.

Regards,
Fabio.

Hi Fabio,

Thanks for the reply but i have one query regarding your answer that question says that "A set of numbers" so there may be infinite numbers since no limit is given of the set. It never says that -1 is the first or lowest number then how can we say that -3 will not be in the set if -1 is there.

jeetu_vishnoi wrote:Hi Fabio,

Thanks for the reply but i have one query regarding your answer that question says that "A set of numbers" so there may be infinite numbers since no limit is given of the set. It never says that -1 is the first or lowest number then how can we say that -3 will not be in the set if -1 is there.

We cannot say -3 will NOT be in the set, but we also cannot guarantee that -3 belongs to the set, that´s the point! I said that -1 could be the lowest term, it is POSSIBLE, just that!!

Regards,
Fábio.

Hey Fabio, Thanks for the explanation to this. I remember seeing this problem in one of the books and I could not figure out why -3 was not correct. The explanation in the book did not address it. Your explanation clears it up. Thanks again...

fskilnik wrote:

jeetu_vishnoi wrote:Hi Fabio,

Thanks for the reply but i have one query regarding your answer that question says that "A set of numbers" so there may be infinite numbers since no limit is given of the set. It never says that -1 is the first or lowest number then how can we say that -3 will not be in the set if -1 is there.

We cannot say -3 will NOT be in the set, but we also cannot guarantee that -3 belongs to the set, that´s the point! I said that -1 could be the lowest term, it is POSSIBLE, just that!!

Regards,
Fábio.

Hi Fabio,

In this case if -1 is the lowest, then -1 could be largest also, so then the remaining terms could be existing or not existing also.. What do you think about this?

rkanthilal wrote:Hey Fabio, Thanks for the explanation to this. I remember seeing this problem in one of the books and I could not figure out why -3 was not correct. The explanation in the book did not address it. Your explanation clears it up. Thanks again...

My pleasure!

allizzwell wrote:Hi Fabio,

In this case if -1 is the lowest, then -1 could be largest also, so then the remaining terms could be existing or not existing also.. What do you think about this?

No, allizzwell!

The question stem says that -1 belongs to the set, and that for EVERY t element that is there (and -1 is certainly one of them), you will have (t+2) also belonging to the set!

That means that if -1 is there, you are SURE there are infinite integer numbers there, please observe that ALL positive odd numbers are certainly there, and we can guarantee that there is no greatest element, FOR SURE! Got it?

Important: I CANNOT say that if -1 is there, then the set is JUST the odd integers from -1 up... this is NOT necessarily true, because perhaps (say) 10 would also belong to the set (why not?), and if so, all even numbers from 10 up would also be in the set, etc etc etc...

fskilnik wrote:

allizzwell wrote:Hi Fabio,

In this case if -1 is the lowest, then -1 could be largest also, so then the remaining terms could be existing or not existing also.. What do you think about this?

No, allizzwell!

The question stem says that -1 belongs to the set, and that for EVERY t element that is there (and -1 is certainly one of them), you will have (t+2) also belonging to the set!

That means that if -1 is there, you are SURE there are infinite integer numbers there, please observe that ALL positive odd numbers are certainly there, and we can guarantee that there is no greatest element, FOR SURE! Got it?

Important: I CANNOT say that if -1 is there, then the set is JUST the odd integers from -1 up... this is NOT necessarily true, because perhaps (say) 10 would also belong to the set (why not?), and if so, all even numbers from 10 up would also be in the set, etc etc etc...

Hi Fabio,

I am sorry but still it is not clear, since we have -1 into the set then it should also satisfy the given condition of (t+2). I mean to say that -1 should also be calculated with the same formula

t+2 = -1 or
t= -3

jeetu_vishnoi wrote: I am sorry but still it is not clear, since we have -1 into the set then it should also satisfy the given condition of (t+2). I mean to say that -1 should also be calculated with the same formula

t+2 = -1 or
t= -3

Hi jeetu, no problem, I will explain to you in a "more formal" way that I guess it will makes it crystal clear...

When we say that A implies B, we CANNOT infer the reverse (that is called the reciprocal statement)... you must pay attention to this in Critical Reasoning questions, by the way!!

In our case, we have

A = t is there
B = t+2 is there

And it is given that A implies B, so that when we know someone is there, two units bigger is there, too. But we cannot use B implies A, because this was NOT given and cannot be inferred!

Look: if you put t+2 = -1 , I agree that t = -3, but YOU believe that if t+2 is there, t must be there... you shouldn´t!! You may say that t = -3 (correct), but that does not mean that -3 must belong to the set... understood?

Let me show you another example:

If x= 2 and y = 3 then x+y = 5, correct?

I did not say that (x+y = 5) means (x=2 and y=3) because this is NOT true, for sure!!

Is that clear now?

(If not, please keep asking, really.)

Regards,
Fabio.