1Setsuna KujoAug 08, 2021Two cases: eitherf(a) < 0, f(b) > 0orf(a) > 0, f(b) < 0Given f(x) is continuous from a to b, it must not jump. And yet it must cross 0 in order to go from <0 to >0 (or the other way around). Therefore it must at some point equal 0.Is this proof valid?
1olive79popSep 13, 2021No, I would restate it as "Many [relatively] easy statements in mathematics are [relatively] hard to prove."
0tapsivofyoJun 24, 2021I would like to recommend you guys to visit What are the advantages and disadvantages of the PDF format. I am sure they will make your work easier than before. Highly recommended to you guys!
Looks to me like a variation of Rolle's theorem.
Two cases: either
f(a) < 0, f(b) > 0
or
f(a) > 0, f(b) < 0
Given f(x) is continuous from a to b, it must not jump. And yet it must cross 0 in order to go from <0 to >0 (or the other way around). Therefore it must at some point equal 0.
Is this proof valid?
I would like to get a more Rigour proof
i'm not the best at writing proofs, sorry
No, I would restate it as "Many [relatively] easy statements in mathematics are [relatively] hard to prove."
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