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Simple proof in analysis (where did I go wrong)

  • Thread starter ych22
  • Start date
  • #1
115
1

Homework Statement



Let a,b be any real number. Show that if a-e < b for any e > 0, then a <= b.

Homework Equations





The Attempt at a Solution



I tried a proof by contradiction, that if a>b then we violate some assumption.
Here it goes:

Suppose not.
Then a > b.
a- e > b - e
Since we assume a-e < b, thus b > b-e. But well...this seems right. Where did I go wrong?
 

Answers and Replies

  • #2
You didn't go wrong anywhere because you didn't do anything. However, I don't think your approach would help you with the proof since if a-e > b-e and b>b-e then a-e >b. There is also the possibility that b>a-e.

I would consider to the cases
0 <a-b<e and
a-b <0 <e.

The second case is sort of trival. The first case is sort of a limit definition. So you can show a=b in the first case and a<b in the second.
 
Last edited:

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