Exercise 2.1.37

Answers

Let $c = \frac{a}{b} \in ℚ$ . We have that

T (x) = T (\underset{b times}{\underset{⏟}{\frac{1}{b} x + \frac{1}{b} x + \dots + \frac{1}{b} x})} = bT (\frac{1}{b} x)

and hence $T (\frac{1}{b} x) = \frac{1}{b} T (x)$ . So finally we have

T (cx) = T (\frac{a}{b} x) = T (\underset{a times}{\underset{⏟}{\frac{1}{b} x + \frac{1}{b} x + \dots + \frac{1}{b} x})}

= aT (\frac{1}{b} x) = \frac{a}{b} T (x) = cT (x) .

jephianlin

2011-06-27 00:00