If I had two numbers $a,b \in \mathbb{R}$, is it true that $$(\forall \epsilon \in \mathbb{R}, \epsilon > 0), |a-b| < \epsilon \implies a = b$$ This is not part of any homework or assignment, but it is something I was wondering about. If it is the case, how could I rigorously prove it?
Asked
Active
Viewed 106 times
1
-
Do you know that the only non-negative real number which is less than every positive real number is $0$? The same statement holds if word "real" is replaced by "rational". Does the corresponding rational version seem obvious. It ultimately boils down to the fact that "Given a natural number there exists a greater one". – Paramanand Singh May 22 '20 at 07:08
1 Answers
1
You can use contradiction. Assume $ a\neq b \implies a -b \neq 0 \implies |a -b|\neq 0 $, then by definition of the absolute value we have that $|a-b| > 0$. Since $\epsilon$ can be any positive real number, in particular let $\epsilon = \frac{|a-b|}{2} >0$. Then, from our hypothesis we get that $$ |a-b| <\frac{|a-b|}{2} \implies 2|a-b| < |a-b| \implies 2 < 1 $$ where we get a contradiction. This means our original assumption was wrong, and hence $ a=b$.
Notes: Notice that there was nothing special about choosing $\epsilon = \frac{|a-b|}{2}$, you can choose $\epsilon = \frac{|a-b|}{n}$ for any $n \in \mathbb{N}\cap[1, \infty)$ and the proof follows the same. Also notice that in the last step we cancel out the $|a-b|$, which we can do because we established previously that $|a -b|\neq 0 $.
Robert Lee
- 7,233