The answer is 1/2, the larger of the individual factors. |
Here is the reason: |
d(T(x, y), T(x', y')) = d((x/2, y/3), (x'/2, y'/3)) |
= ((x/2 - x'/2)2 + (y/3 - y'/3)2)1/2 |
= ((1/4)⋅(x - x')2 + (1/9)⋅(y - y')2)1/2 |
≤ ((1/4)⋅(x - x')2 + (1/4)⋅(y - y')2)1/2 (Do you see why?) |
= (1/2)⋅((x - x')2 + (y - y')2)1/2 |
= (1/2)⋅d((x, y), (x', y')) |
To show no number smaller than 1/2 works for all points, take y = y' = 0. Then |
d(T(x, y), T(x', y')) = d((x/2, 0), (x'/2, 0)) |
= ((x/2 - x'/2)2)1/2 |
= ((1/4)⋅(x - x')2)1/2 |
= (1/2)⋅((x - x')2)1/2 |
= (1/2)⋅d((x, y), (x', y')). |
Return to Contractions