Here is the theorem from Steven Abbot's Understanding Analysis.

Theorem. Two real numbers $a$ and $b$ are equal if and only if for every real number $\epsilon > 0$ it follows that $|a - b| < \epsilon$.

I have a two part question.

  1. How do I write the theorem using quantifiers?

  2. How is contradiction being used with the quantifiers? Can you show with the quantified statement how we are doing proof by contradiction?

Here's an image of the page:

Page from Abbot's "Understanding Analysis".

  • $\begingroup$ The theorem is written with quantifiers! The phrases "for all" and "there exists" are quantifiers. $\endgroup$ Apr 23 at 20:06

1 Answer 1


It is already written using quantifiers "for all" and "there exists" but if you want $$\forall a,b:\,\Big(a=b\iff(\forall\varepsilon>0:\,|a-b|<\varepsilon)\Big)$$ The direction $$\forall a,b:\,\Big(a=b\implies(\forall\varepsilon>0:\,|a-b|<\varepsilon)\Big)$$ is called trivial and the direction $$\forall a,b:\,\Big((\forall\varepsilon>0:\,|a-b|<\varepsilon)\implies a=b\Big)$$ is done by proving that the negation $$\exists a,b:\,\Big((\forall\varepsilon>0:\,|a-b|<\varepsilon)\ \mathrm{and}\ a\neq b\Big)$$ leads to a contradiction.

Hope this helps. :)

  • $\begingroup$ This does help a lot! What about a, and b? Wouldn’t they have the quantifier for all. Where would that go and how does that factor into the contradiction proof? Thank you! $\endgroup$
    – Dr. J
    Apr 23 at 20:25
  • $\begingroup$ Yes, $a$ and $b$ do need quantifiers. The "for all" quantifier would go right in front of them. $\endgroup$
    – Lee Mosher
    Apr 23 at 20:27
  • $\begingroup$ @Dr.J I edited my answer. $\endgroup$ Apr 23 at 20:29
  • $\begingroup$ Thank you! This makes sense. I'm struggling to understand contradiction and the contrapositive with quantifiers. For example, you negated the for all quantifier for a, and b. I was thinking it would stay the same. Why do we negate it? Also would the contrapositive for the 3rd formula you provided be this: $$\forall a,b:\,\Big(a \neq b\implies(\exists\varepsilon>0:\,|a-b|>\varepsilon)\Big)$$ $\endgroup$
    – Dr. J
    Apr 23 at 20:51
  • $\begingroup$ @Dr.J: Think about it: you have the statement "property $P$ holds for all $x$". Is the negation "property $P$ does not hold for all $x$ or "property $P$ does not hold for at least one $x$? For your contrapositive, it is correct other than the minor $\geq\varepsilon$ instead of $>\varepsilon$. $\endgroup$ Apr 23 at 20:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .