Show |x-a|< epsilon IFF a-epsilon < x < a+epsilon

[Kinetica]

Homework Statement

Hi! I am new to Real Analysis. Please let me know if my solution is alright. Thanks.
epsilon=e

Show that |x-a|< e IFF a-e < x < a+e

The Attempt at a Solution

Assume |x-a|< e. Prove, a-e < x < a+e

|x-a|< e
-e < x-a < e
a-e < x < a+e

Assume a-e < x < a+e. Prove |x-a|< e.

a-e < x < a+e
-e < x-a < e
|x-a|< e

 

[hunt_mat]

Looks okay to me.

 

[Coto]

Looks good to me to, but just to improve the style of your proof: when an operation is reversible, you can use the \iff sign and combine your proofs from each direction into a single proof.

E.g:
\|x-a\| &lt; \epsilon \iff -\epsilon &lt; x - a &lt; \epsilon \iff a - \epsilon &lt; x &lt; a + \epsilon

 

[Kinetica]

Thank you, this is a great suggestion. My proofs are indeed full of organizational flaws.