Irrational inequalities √f(x)>g(x) and √f(x)>g(x)

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Discussion Overview

The discussion revolves around the inequalities involving the square root of a function, specifically the conditions under which the inequalities √f(x) < g(x) and √f(x) > g(x) hold true. Participants explore the implications of these inequalities in terms of the functions f(x) and g(x), including their non-negativity and the relationships between them.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the inequality √f(x) < g(x) is equivalent to f(x) ≥ 0, g(x) > 0, and f(x) < (g(x))^2, questioning why g(x) must be strictly greater than zero.
  • Others argue that g(x) can be equal to zero in certain contexts, leading to confusion about the equivalence of the inequalities.
  • A participant suggests that if √f(x) > g(x), then f(x) can equal zero, but this would contradict the requirement that g(x) is greater than zero.
  • There is a discussion about the implications of using g(x) ≥ 0 versus g(x) > 0 in the context of the inequalities, with some asserting that both can be used interchangeably under certain conditions.
  • Several participants express uncertainty about the implications of the inequalities when the conditions change to include ≤ or ≥, leading to further questions about the relationships between f(x) and g(x).
  • Disagreements arise regarding whether the conditions for the inequalities should include strict inequalities or if non-strict inequalities can be applied without issue.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the conditions required for the inequalities involving √f(x) and g(x). There are multiple competing views regarding the necessity of g(x) being strictly greater than zero and the implications of using different forms of the inequalities.

Contextual Notes

Some participants note that quantifiers such as "for all real x" are missing from the discussion, which may affect the interpretation of the inequalities. Additionally, there is uncertainty about the implications of changing the inequality signs from < to ≤ or > to ≥.

Danijel
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So, I know that the inequality √f(x)<g(x) is equivalent to f(x)≥0 ∧ g(x)> 0 ∧ f(x)<(g(x))^2. However, why does g(x) have to be greater and not greater or equal to zero? Is it because for some x, f(x) = g(x)=0, and then > wouldn't hold? Doesn't f(x)<(g(x))^2 make sure that f(x) will not be equal to g(x)?
If that is so, then how is now √f(x) < g(x) equivalent to f(x)≥0 ∧ g(x)≥0 ∧ f(x)>(g(x))^2, where g(x) can now be equal to zero?
Also, what happens to the conditions when √f(x)≤g(x) or √f(x)≥g(x)?
 
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There are some quantifiers missing, like "for all real x".
Danijel said:
Doesn't f(x)<(g(x))^2 make sure that f(x) will not be equal to g(x)?
It does. That doesn't make the requirement g(x)> 0 wrong, it just makes it unnecessary.
Danijel said:
where g(x) can now be equal to zero?
It cannot be equal to zero.
Danijel said:
Also, what happens to the conditions when √f(x)≤g(x) or √f(x)≥g(x)?
What do you think?
 
mfb said:
There are some quantifiers missing, like "for all real x".It does. That doesn't make the requirement g(x)> 0 wrong, it just makes it unnecessary.It cannot be equal to zero.What do you think?
Does that mean that √f(x) < g(x) is not equivalent to f(x)≥0 ∧ g(x)≥0 ∧ f(x)>(g(x))^2, but to f(x)≥0 ∧ g(x)>0 ∧ f(x)>(g(x))^2, or both, since f(x)>(g(x))^2 implies that g(x) cannot be equal to zero, so it is the same thing? I don't understand then why would anybody write g>0 for √f(x)<g(x) and and g≥0 for √f(x)>g(x), if g≥0 or g>0 can be written in both. It is confusing.
For your question, I think that everything stays the same, except the part f(x)≤(g(x))^2, where < is now ≤.
Thank you for your reply.
 
mfb said:
There are some quantifiers missing, like "for all real x".It does. That doesn't make the requirement g(x)> 0 wrong, it just makes it unnecessary.It cannot be equal to zero.What do you think?
Also, if √f(x)> g(x), and if f(x)≥0 and g(x)>0 and f(x) > (g(x))^2 then f(x) can = 0 , but 0 is not greater than any g(x) > 0, so f(x) should be strictly greater than zero.
 
I meant that √f(x) > g(x) is equivalent to f(x)≥0 ∧ g(x)≥0 ∧ f(x)>(g(x))^2. Sorry.
 
Danijel said:
Does that mean that √f(x) < g(x) is not equivalent to f(x)≥0 ∧ g(x)≥0 ∧ f(x)>(g(x))^2, but to f(x)≥0 ∧ g(x)>0 ∧ f(x)>(g(x))^2, or both, since f(x)>(g(x))^2 implies that g(x) cannot be equal to zero, so it is the same thing?
The two are the same thing, right. The second one makes it easier to see that g(x) cannot be zero.
Danijel said:
and and g≥0 for √f(x)>g(x)
Careful, this is a completely different expression.
Danijel said:
Also, if √f(x)> g(x), and if f(x)≥0 and g(x)>0 and f(x) > (g(x))^2 then f(x) can = 0 , but 0 is not greater than any g(x) > 0, so f(x) should be strictly greater than zero.
You used a wrong conclusion to derive another wrong conclusion. If you have √f(x)>g(x), why should g(x) be larger than 0? What is wrong with negative g(x)?
 
mfb said:
The two are the same thing, right. The second one makes it easier to see that g(x) cannot be zero.Careful, this is a completely different expression.You used a wrong conclusion to derive another wrong conclusion. If you have √f(x)>g(x), why should g(x) be larger than 0? What is wrong with negative g(x)?
I used wrong to derive wrong, intending to show that the former had to be wrong. That is, I tried to find a contradiction.
 
mfb said:
The two are the same thing, right. The second one makes it easier to see that g(x) cannot be zero.Careful, this is a completely different expression.You used a wrong conclusion to derive another wrong conclusion. If you have √f(x)>g(x), why should g(x) be larger than 0? What is wrong with negative g(x)?
There is nothing wrong, sorry. I overlooked it.
 

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