Originally Posted by

**ryu1** I read it couple of times and I cant understand this part, can you explain it in a simpler fashion please? like how would you write it for real (like some parts you wrote at previous sections of your post i.e ""Towards contradiction, assume that there exists some t such that h(t) ≤ 0. If h(t) = 0, then f(t) = t, which contradicts the assumption that f(x) ≠ x for all x". ")

Need to prove f(x) > x, for all x.

f(x) > x = f(x) - x > 0

Defining a new function h(x) = f(x) - x.

Now we need to prove h(x) > 0 for all x.

f(1) = 2 , given, therefore h(1) = 1 > 0

Let's assume towards contradiction there is some t such that h(t) <= 0, if h(t) = 0 then h(t) = f(t) - t = 0 , f(t) = t which contradicts the condition f(x) != x.

Now we still need to prove towards contradiction there is some t such that h(t) < 0.

According to the IVT, in the closed segment [1,t] there is some x such that h(x) = 0, because h(1) > 0 and h(t) < 0.

That x also contradicts the condition f(x) != x (again because h(x) = 0 = f(x) - x = 0 = f(x) = x).

Thus we get a contradiction for h(t) <= 0, which means h(x) > 0 for all x, which proves f(x) > x for all x.