Is the Following Proof Correct?

Proposition. Let $a, b$ be rational numbers. Show that $a = b$ if and only if $\text{LIM}_{n\to\infty}a = \text{LIM}_{n\to\infty}b$ (i.e., the Cauchy sequences $a, a, a, a,\dots$ and $b, b, b, b,\dots$ are equivalent if and only if $a = b$). This allows us to embed the rational numbers inside the real numbers in a well-defined manner.

Proof. Assume $a=b$ and let $\epsilon>0$ and $N$ be any natural number, evidently given any $n\ge N$ we have $|a_n-b_n| = |a-b| = 0\leq\epsilon$ implying that $(a)_{n=0}^{\infty}$ and $(b)_{n=0}^{\infty}$ are equivalent sequences.

Conversely let $(a)_{n=0}^{\infty}$ and $ (b)_{n=0}^{\infty}$ be equivalent sequences consequently given any $\epsilon>0$ for some $N\in\mathbf{N}$ we have $\forall n\ge N(|a_n-b_n| = |a-b|\leq\epsilon)$. Proposition $\textbf{4.2.9}$ establishes that exactly one of $a<b$, $a>b$ or $a=b$ may be true.

Assume that $a<b$ consequently $(b-a) = x\in\mathbf{Q}^+$ then by hypothesis $|a_N-b_N| = |b_N-a_N| = |x|\leq\frac{|x|}{2}$ a contradiction. We may infer a similar contradiction by assuming $a>b$ consequently it must be that $a=b$.


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    $\begingroup$ It seems fine for me $\endgroup$ – Stefan4024 Jun 19 '18 at 23:18
  • $\begingroup$ You can use \lim instead of \text{LIM}. $\endgroup$ – qwr Jun 20 '18 at 1:11

YES, your proof is correct.

And here's an easier way to prove the statement:

$\lim_{n \to \infty} a = \lim_{n \to \infty} b$

$\implies \lim_{n \to \infty} (a \cdot 1) = \lim_{n \to \infty} (b \cdot 1)$

$\implies a \cdot \lim_{n \to \infty} 1 = b \cdot \lim_{n \to \infty} 1$

$\implies a \cdot 1 = b \cdot 1$

$\implies a=b.$

  • $\begingroup$ Thankyou for your response but that does not quite answer my question. $\endgroup$ – Atif Farooq Jun 20 '18 at 0:54
  • $\begingroup$ Sorry, I edited the answer. $\endgroup$ – Twink Jun 20 '18 at 1:01

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