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In the book "Mathematical analysis and it's nature" in order to prove the uniqueness of some quantities it would assume that it is not unique (as usual). To prove that the supposed $x$ and $y$ are equal therefore the quantity in question is unique it proves the following: $$\text{given $\varepsilon \gt 0$}$$ $$ \vert x-y\vert \lt \varepsilon$$ $$\therefore x=y $$ I cannot understand why would x be equal to y when the said condition is true. it is written in the book that "Because $0$ is the only non-negative real number that is less than every positive real number" but that won't help me to comprehend either.

Hannan
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    Check this: https://math.stackexchange.com/q/3436553/42969 – Martin R Nov 29 '23 at 09:21
  • I say "give me a small number $\epsilon$" and you say $10^{-10}$. I say the difference is smaller than that, give me another number. You say $10^{-50}$ and I say the difference is smaller than that. You keep going and no matter how small you make your number, the difference is always smaller. What do you conclude? – Paul Nov 29 '23 at 09:25
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    This question made me this of this very relevant and active question on matheducators.se: https://matheducators.stackexchange.com/questions/26934/to-17-year-olds-how-can-i-explain-that-two-numbers-with-arbitrarily-small-diffe – Adam Rubinson Nov 29 '23 at 10:49

3 Answers3

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Alternative approach:

Suppose $~x \neq y.$

Then $~x - y \neq 0.$

Then $~|x - y| > 0.~$

Choose $~\epsilon = \dfrac{|x - y|}{2}.~$

This violates the premise that for all $~\epsilon > 0,~$

that $~|x - y| < \epsilon.~$

Therefore, the assumption that $~x \neq y~$ has led to a contradiction.

Therefore, $~x = y.$

user2661923
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If $0$ is the only non-negative real number that is less than every positive real number and you have that for every $\varepsilon >0$ (i.e. for every positive real number) $|x-y|<\varepsilon$ (i.e. $|x-y|$ is less than every positive real number), then it means that $|x-y|$ is indeed $0$.

Now $|x-y|=\pm (x-y)$ and both $x-y=0$ and $y-x=0$ imply $x=y$.

SilvioM
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It is beneficial to think about the term "given $\epsilon > 0$" it means the same as "for every" (a.k.a. $\forall$) $\epsilon > 0$. And yep this approach is a bit philosophical, (see Zenon's paradox about the turtle and Achilles), but let's see this: Let $a = |x-y|$, our first claim is that $a$ is a nonnegative number. On the other hand it is smaller than any positive number, since for each $\epsilon > 0$ if we get an $\epsilon$ then $a < \epsilon$. If we assume that $a > 0$ it contradicts the assumption that it is smaller than any positive number therefore $a=0=|x-y|\implies x = y.$

Koppány Dodony
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