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

[Danijel]

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)?

 

[mfb]

There are some quantifiers missing, like "for all real x".

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.

where g(x) can now be equal to zero?

It cannot be equal to zero.

Also, what happens to the conditions when √f(x)≤g(x) or √f(x)≥g(x)?

What do you think?

 

[Danijel]

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.

 

[Danijel]

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.

 

[Danijel]

I meant that √f(x) > g(x) is equivalent to f(x)≥0 ∧ g(x)≥0 ∧ f(x)>(g(x))^2. Sorry.

 

[mfb]

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.

and and g≥0 for √f(x)>g(x)

Careful, this is a completely different expression.

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)?

 

[Danijel]

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.

 

[Danijel]

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.