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 Tathiana Garbin
Émilie Loit
Runner-up Dája Bedáňová
Dinara Safina
Final score6–3, 3–6, 6–4

Tathiana Garbin and Émilie Loit won the title.[1][2][3]

Seeds

  1. Virginia Ruano Pascual / Magüi Serna (first round)
  2. Tathiana Garbin / Émilie Loit (winners)
  3. Eugenia Kulikovskaia / Tatiana Poutchek (semifinals)
  4. Catherine Barclay / Martina Müller (first round)

Results

Key

Draw

First Round Quarterfinals Semifinals Final
1 V Ruano Pascual
M Serna
3 64
J Kostanić
M Matevžič
6 7 J Kostanić
M Matevžič
2 3
C Martínez Granados
F Pennetta
3 2 D Bedáňová
D Safina
6 6
D Bedáňová
D Safina
6 6 D Bedáňová
D Safina
6 2 7
3 E Kulikovskaia
T Poutchek
6 6 3 E Kulikovskaia
T Poutchek
3 6 5
S Stone
S Stosur
4 2 3 E Kulikovskaia
T Poutchek
2 6 6
Y-J Cho
Ad Serra Zanetti
3 6 4 Q M Bartoli
S Foretz
6 4 2
Q M Bartoli
S Foretz
6 3 6 D Bedáňová
D Safina
3 6 4
WC C Dellacqua
N Sewell
5 5 2 T Garbin
É Loit
6 3 6
L Granville
M Tu
7 7 L Granville
M Tu
3 6 1
R McQuillan
L McShea
6 7 R McQuillan
L McShea
6 4 6
4 C Barclay
M Müller
1 5 R McQuillan
L McShea
4 1
V Razzano
S Talaja
62 7 3 2 T Garbin
É Loit
6 6
C Fernández
A Widjaja
7 66 6 C Fernández
A Widjaja
6 2 65
An Barna
M Weingärtner
62 3 2 T Garbin
É Loit
3 6 7
2 T Garbin
É Loit
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.

References

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