Proving 3/4 is Infimum of A: A = { x^2 +x + 1 }

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SUMMARY

The discussion centers on proving that 3/4 is the infimum of the set A = { x^2 + x + 1 }. Participants emphasize the need for a formal proof and clarify the conditions under which the proof holds, specifically noting that x must be a real number (x ∈ R). The contradiction arises when assuming a value M < 3/4, leading to the conclusion that x^2 + x + 1 must exceed M, thus confirming 3/4 as the infimum.

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mercuryman
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Homework Statement



Giva a formal proof that 3/4 is the infimum of the set : A = { x^2 +x + 1 }

Homework Equations


I need a clear way to prove it - to understand the contradiction.


The Attempt at a Solution


I've assumed there is a M<3/4 that x^2 +x + 1>M. where is the contradiction?
 
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I've assumed there is a M<3/4 that x^2 +x + 1>M
I think you mean ##x^2 +x + 1\leq M##? Otherwise it would be a bit pointless.

You should have given ##x \in R## somewhere.
 
Why is this posted under "physics"? It is clearly a math problem- complete the square.
 

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