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-1 votes
2 answers
145 views

prove quadratic polynomial has no real roots

The problem asks me to prove that a polynomial $f(x)=x^2+ax+b$ has no real roots for some $a,b \in \Bbb{R}$ I started by assuming that $f(x)=x^2+ax+b$ has real roots and therefore the determinant $a^...
Skrrrrrtttt's user avatar
-1 votes
1 answer
54 views

Prove that there exist equal number of irrational numbers between any 2 rational numbers, when the difference between the 2 rational numbers is same. [closed]

Prove that there exist equal number of irrational numbers between any 2 rational numbers, when the difference between the 2 rational numbers is same. If the assertion is not true then please prove ...
Ash_Tag's user avatar
-1 votes
1 answer
82 views

Polynomial proof in Real numbers [closed]

how to prove that there exists a 2 variable polynomial which is bounded below and the range of values is strict subset from the $\mathbb{R}$.
Albert's user avatar
  • 3
-2 votes
2 answers
48 views

A natural number between two reals [closed]

How should I go about proving the following: $\forall x \in \mathbb{R}, \exists n \in \mathbb{N}$ $ s.t. $ $20(3x^2 - 3x + 2) > 15n > 12(5x^2 - 5x + 2)$
Zexion12's user avatar
  • 319
-3 votes
4 answers
109 views

why is the co-prime part not mentioned in the definition of the rational number?

Proving $\sqrt{2}$ an irrational number is a quite popular exercise, in precalculus courses, but if we look clearly the definition that is introduced, in the beginning of the course, it never ...
Yanjan. Kaf.'s user avatar
-3 votes
1 answer
67 views

Proving: $a^2 < b^2 ⇔ |a| < |b|$

I started studying mechanical engineering and it works perfectly fine for me but i stumbled across this problem: $$a^2 < b^2 ⇔ |a| < |b|$$ I found a solution but that took me a full piece of ...
Finn Eggers's user avatar
-3 votes
1 answer
75 views

How can I prove this question?

We know that it is not proved that $e^e$ is transcendental, so neither is the number that $e^{e\sqrt{2}}$. My question is, if one turns out to be, how can it be proved that the other is? Because there ...
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