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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.

Siong Thye Goh
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futuremathteach
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  • The set is all the numbers that have an integer square root. In other words, it is the set of perfect squares. So you need to prove that the perfect squares are unbounded. – Jaap Scherphuis Jan 24 '20 at 09:02

3 Answers3

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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.
Siong Thye Goh
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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.

Peter Szilas
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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

subzero
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