2013 Fifth Third Bank Tennis Championships – Men's Doubles
Austin Krajicek and John Peers were the defending champions, but Peers chose not to compete.
Krajicek paired with Mitchell Krueger but lost in the semifinals to eventual finalists Bradley Klahn and Michael Venus.
Frank Dancevic and Peter Polansky won the title 7–5, 6–3.
Men's Doubles | |
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2013 Fifth Third Bank Tennis Championships | |
Champions | ![]() ![]() |
Runners-up | ![]() ![]() |
Final score | 7–5, 6–3 |
Seeds
Purav Raja / Divij Sharan (First Round) Adam Feeney / Matt Reid (First Round) Maxime Authom / Ruben Bemelmans (Semifinals) Bradley Klahn / Michael Venus (Final)
Draw
Key
- Q = Qualifier
- WC = Wild Card
- LL = Lucky Loser
- Alt = Alternate
- SE = Special Exempt
- PR = Protected Ranking
- ITF = ITF entry
- JE = Junior Exempt
- w/o = Walkover
- r = Retired
- d = Defaulted
First round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | ![]() ![]() | 4 | 710 | [5] | |||||||||||||||||||||||
![]() ![]() | 6 | 68 | [10] | ![]() ![]() | 4 | 4 | |||||||||||||||||||||
WC | ![]() ![]() | 3 | 65 | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
![]() ![]() | 6 | 77 | ![]() ![]() | 4 | 5 | ||||||||||||||||||||||
4 | ![]() ![]() | 77 | 6 | 4 | ![]() ![]() | 6 | 7 | ||||||||||||||||||||
![]() ![]() | 64 | 2 | 4 | ![]() ![]() | 7 | 6 | |||||||||||||||||||||
![]() ![]() | 7 | 6 | ![]() ![]() | 5 | 3 | ||||||||||||||||||||||
WC | ![]() ![]() | 5 | 2 | 4 | ![]() ![]() | 5 | 3 | ||||||||||||||||||||
![]() ![]() | w/o | ![]() ![]() | 7 | 6 | |||||||||||||||||||||||
WC | ![]() ![]() | ![]() ![]() | 4 | 2 | |||||||||||||||||||||||
![]() ![]() | 2 | 1 | 3 | ![]() ![]() | 6 | 6 | |||||||||||||||||||||
3 | ![]() ![]() | 6 | 6 | 3 | ![]() ![]() | 4 | 6 | [4] | |||||||||||||||||||
![]() ![]() | 6 | 6 | ![]() ![]() | 6 | 4 | [10] | |||||||||||||||||||||
![]() ![]() | 2 | 2 | ![]() ![]() | 6 | 6 | ||||||||||||||||||||||
![]() ![]() | 7 | 1 | [10] | ![]() ![]() | 3 | 4 | |||||||||||||||||||||
2 | ![]() ![]() | 5 | 6 | [7] |
gollark: However, apiohax = P = (maybe) NP = 0 (mod N). Therefore, as rings may be noncommutative, it is the case that the left ideal, 7, is an eigenvalue of the matrix expansion of the general bee formula. By basic applications of previously proven lemmas, it can be shown that this makes apiohax isomorphic to the group (ℤ, +). The implications are obvious.
gollark: If you mean the history, this is now classified.
gollark: ?urban apioform
gollark: As apiohax exists and is known to, you exist.
gollark: However, apiohax existing does not contradict apiohax existing.
References
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