Proof that $\sqrt{x}$ is Unbounded!

Question

I need help making my proof strong/valid! I am new to proofs.

The question was:Let $A=\{x\in \mathbb{N} | \sqrt{x}\in \mathbb{N}\}$. Prove that A is not bounded.

Proof: We must show that for all $m\in \mathbb{N}$, we can find an $x_0 \in B$ such that |$x_0| > M$. Suppose we are given $M \in \mathbb{N}$. By the definition of Archimedean, there exists an $N \in \mathbb{N}$ where $N > M-1$. $ \sqrt{N} \in B,$ let $x_0= \sqrt{N} > \sqrt{M-1} > M$. Since M was arbitrary, we can conclude that B was unbounded.

Answer 1

We must show that for all $M\in \mathbb{N}$, we can find an $|x_0| \in A$ such that $x_0 > M$.

Suppose we are given $M \in \mathbb{N}$. Let $x_0 = (M+1)^2 \in \mathbb{N}$, then $\sqrt{x}=M+1 > M$.

Remark about your attempt:

• notation wise, I believe $A$ and $B$ are the same set. Avoiding using new notation without defining it.
• I don't see the rational of $\sqrt{N} \in A$. In particular, I don't see why square root of an arbitary natural number is promised to be a natural number.
• $\sqrt{M-1} >M$? I don't see why this is true.

Answer 2

Consider the sequence:

1)$a_n = √n$, $n\in \mathbb{N}$, and the

subsequence :

2) $a_{n^2} = n$;

Archimedes:

For $M \gt 0,$ $M$ real, there exists a $n_0$

with. $n_0> M.$.

Thus: $a_{n_0^2} = n_0 >M.$

Since $f(x)=√x$ is an increasing function:

For $n \ge (n_0)^2 $: $a_n \gt M$,

hence unbounded.

Answer 3

if {a} is a sequence an = n^1/2. if a < M where M(>0)is a upper bound of a. then n^1/2 < M ==> n < M^2. which restricts the terms of the sequence hence a