set theory maths please help

Question

$\Bbb N$ is the set of natural numbers $\Bbb N=\{0,1,2,\dots\}$ . For every $n\in\Bbb N$, let $A_n = \{ x\in \Bbb N \,\vert\, 0\leq x \leq n\}$.

Prove or Disprove the following:

$$\forall_{n \in \Bbb N}, \forall_{m \in \Bbb N}, (A_m = \{x^2 \,|\, x\in A_n\}) \iff (m=n \wedge n\lt2)$$

I tried it two times, got two different answers

first time: my answer is, and I'm not sure. The right sides intersection of n with then $n<2$, means there are the only options of $0$ or $1$, which makes the left side wrong.
second time, made me think that its true, since both $0^2= 0$ and $1^2 = 1$

not sure which one is right?

This is hard to read because of the lack of Mathjax. Have you considered learning how to typeset mathematics? — Edward Evans, Mar 27 '16 at 16:49
If I am reading correctly, the result you are trying to prove or disprove is correct. For if $m\ge 2$ then $A_m$ contains some non-squares. You probably mean for every $n\in \mathbb{N}$ when defining $A_n$. — André Nicolas, Mar 27 '16 at 17:07
Since $2$ is the smallest integer that is not a square, it's relatively obvious that you need $n<2$ (otherwise there is a "hole" in ${k^2,|k\in A_n}$). — Jean-Claude Arbaut, Mar 27 '16 at 17:27

score 0 · Answer 1 · answered Mar 27 '16 at 17:37

0

$m=n$ and $n < 2$ is equivalent to "either $m=n=0$ or $m=n=1$".

Let $P$ be the statement that $A_m = \{x^2 : x \in A_n\}$. Then $P$ is equivalent to "$\{ k \in \mathbb{N} : 0 \leq k \leq n \} = \{ x^2 : 0 \leq k \leq m \}$".

The cardinality of the left-hand side is $n$, and the right-hand side $m$, so if $P$ holds then $n=m$; and if $P$ holds then $\{ x^2 : 0 \leq k \leq n \} = \{ x : 0 \leq k \leq n \}$. If $n > 2$ then $3$ is in the left-hand set but not the right-hand set, so it also implies $n < 2$.

The converse is easy (I claim).

answered Mar 27 '16 at 17:37

Patrick Stevens

36,135

what do you mean by if p holds? – sivsi Mar 27 '16 at 17:44
please do not edit any more, this is just right thank you – sivsi Mar 27 '16 at 17:45
@sivsi "If $P$ is true, then…" – Patrick Stevens Mar 27 '16 at 17:45
so therefore it is true? – sivsi Mar 27 '16 at 17:51
Yes, it is true. – Patrick Stevens Mar 27 '16 at 17:55
wow, its much more clear now thank you so much! – sivsi Mar 27 '16 at 18:07

set theory maths please help

1 Answers1