If both non-zero real numbers a and B have f (a + b) = f (a) * f (b) and if x 1, we prove that f (x) is a decreasing function
1 syndrome: make a > 0
∵f(a+0)=f(a)f(0)
∴f(0)=1=f(a-a)=f(a)f(-a)
∵f(-a)>1
∴0
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