Simple proof in analysis (where did I go wrong)

  • Thread starter Thread starter ych22
  • Start date Start date
  • Tags Tags
    Analysis Proof
Click For Summary
SUMMARY

The discussion centers on proving that if \( a - e < b \) for any \( e > 0 \), then \( a \leq b \). The initial proof attempt by contradiction incorrectly assumes \( a > b \) and leads to a valid but misleading conclusion. Participants clarify that the proof requires consideration of two cases: \( 0 < a - b < e \) and \( a - b < 0 < e \). The first case demonstrates \( a = b \), while the second case confirms \( a < b \).

PREREQUISITES
  • Understanding of real number properties
  • Familiarity with proof by contradiction
  • Knowledge of limits and inequalities
  • Basic mathematical logic and reasoning
NEXT STEPS
  • Study the concept of limits in real analysis
  • Learn about proof techniques, specifically proof by contradiction
  • Explore inequalities and their properties in mathematics
  • Review case analysis in mathematical proofs
USEFUL FOR

Students of mathematics, particularly those studying real analysis, educators teaching proof techniques, and anyone interested in enhancing their logical reasoning skills in mathematical contexts.

ych22
Messages
114
Reaction score
0

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?
 
Physics news on Phys.org
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:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K