Answer:
The only thing I can come up with is this:
x<y
⟺x⋅xn−1<y⋅xn−1
Surely y⋅xn−1<y⋅yn−1 because x<y
Then x⋅xn−1<y⋅xn−1<y⋅yn−1
xn<y⋅xn−1<yn, so xn<yn
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Answers & Comments
Answer:
The only thing I can come up with is this:
x<y
⟺x⋅xn−1<y⋅xn−1
Surely y⋅xn−1<y⋅yn−1 because x<y
Then x⋅xn−1<y⋅xn−1<y⋅yn−1
xn<y⋅xn−1<yn, so xn<yn