Direct Proofs: Are They Just Introductions?

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Direct proofs are not merely introductory tools; they are foundational in mathematics and are used to establish significant theorems, including the fundamental theorems of calculus. Many important results in various fields rely on direct proofs to demonstrate relationships between propositions. The discussion highlights that while direct proofs may be introduced in early coursework, their application extends to advanced mathematical concepts. The fundamental theorem of calculus serves as a prime example of a critical theorem derived through direct proof methods. Understanding direct proofs is essential for grasping more complex mathematical theories.
Benn
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Hey guys,

I'm in a proof class right now. We've covered direct proofs and moved on, but I'm still curious about them. Is there any important theorem that has even been derived using a direct proof (assume p to show q) or are they mainly just used to introduce proofs? In class, we only ever cover proofs such as "if n ##\equiv## 1 (mod 2), then n2 ##\equiv## 1 (mod 8)." and the like.

Sorry, I can't get the tex to work out... aha, just got it working, nevermind
 
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Benn said:
Is there any important theorem that has even been derived using a direct proof (assume p to show q) or are they mainly just used to introduce proofs?

Plenty of important theorems are proved using methods of direct proof. I presume that you are familiar with the fundamental theorems of calculus. The standard proofs of these results are done via direct proof. See here: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Proof_of_the_first_part
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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