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Wasi Husain
Jun 22, 2021
  ·  Edited: Jun 22, 2021

Easy Things are hard to prove...

in Video Ideas

Given f : R → R is continuous on [a,b] and sgn( f(a) )sgn( f(b) )= -1

Show that ∃ x∈[a,b] : f(x) = 0


4 answers2 replies
0
2
Ian Fowler
Jun 22, 2021

Looks to me like a variation of Rolle's theorem.

1
Setsuna Kujo
Aug 08, 2021

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?

Wasi Husain
Sep 15, 2021

I would like to get a more Rigour proof

0
Setsuna Kujo
Sep 15, 2021

i'm not the best at writing proofs, sorry

0
1
olive79pop
Sep 13, 2021

No, I would restate it as "Many [relatively] easy statements in mathematics are [relatively] hard to prove."

0
tapsivofyo
Jun 24, 2021

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6 comments