2003 Canberra Women's Classic – Doubles
Nannie de Villiers and Irina Selyutina were the defending champions, but Selyutina chose not to compete, and de Villiers chose to participate in the Hobart International instead.
Doubles | |
---|---|
2003 Canberra Women's Classic | |
Champion | |
Runner-up | |
Final score | 6–3, 3–6, 6–4 |
Tathiana Garbin and Émilie Loit won the title.[1][2][3]
Seeds
Virginia Ruano Pascual / Magüi Serna (first round) Tathiana Garbin / Émilie Loit (winners) Eugenia Kulikovskaia / Tatiana Poutchek (semifinals) Catherine Barclay / Martina Müller (first round)
Results
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
Draw
First Round | Quarterfinals | Semifinals | Final | ||||||||||||||||||||||||
1 | 3 | 64 | |||||||||||||||||||||||||
6 | 7 | 2 | 3 | ||||||||||||||||||||||||
3 | 2 | 6 | 6 | ||||||||||||||||||||||||
6 | 6 | 6 | 2 | 7 | |||||||||||||||||||||||
3 | 6 | 6 | 3 | 3 | 6 | 5 | |||||||||||||||||||||
4 | 2 | 3 | 2 | 6 | 6 | ||||||||||||||||||||||
3 | 6 | 4 | Q | 6 | 4 | 2 | |||||||||||||||||||||
Q | 6 | 3 | 6 | 3 | 6 | 4 | |||||||||||||||||||||
WC | 5 | 5 | 2 | 6 | 3 | 6 | |||||||||||||||||||||
7 | 7 | 3 | 6 | 1 | |||||||||||||||||||||||
6 | 7 | 6 | 4 | 6 | |||||||||||||||||||||||
4 | 1 | 5 | 4 | 1 | |||||||||||||||||||||||
62 | 7 | 3 | 2 | 6 | 6 | ||||||||||||||||||||||
7 | 66 | 6 | 6 | 2 | 65 | ||||||||||||||||||||||
62 | 3 | 2 | 3 | 6 | 7 | ||||||||||||||||||||||
2 | 7 | 6 |
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gollark: Assume there are integers x, y satisfying x²-y²=2(x-y)(x+y)=2x-y, x+y are both integers because they are a sum/difference of integersx, y >= 0 because (-x)²=x² so just ignore negative solutions since they only exist if a positive one does2 has the factors 2,1 so x-y, x+y must be 1, 2 in some orderx-y, x+y differ by 2yx-y, x+y differ by 12y=1y=½But y is an integer
gollark: We are having such advanced intellectual conversations.
gollark: yes.
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