Disproof of an existential proposition

Disproof of an existential proposition

Postby gio_osh » Wed Mar 24, 2021 1:49 am

I would like to disproof:
[tex]\exists x \in N : x > 4 \: \land \: 3x \geq 2^x[/tex].

So far I've got:
[tex]x > 4 \: \rightarrow \: 2^x > 16 \quad[/tex] and [tex]\quad x > 4 \: \leftrightarrow \: 3x > 12[/tex].
But I am stuck on what to do next. Any suggestion would be appreciated.
gio_osh
 
Posts: 2
Joined: Wed Mar 24, 2021 12:46 am
Reputation: 0

Re: Disproof of an existential proposition

Postby romsek » Wed Mar 24, 2021 9:30 am

gio_osh wrote:I would like to disproof:
[tex]\exists x \in N : x > 4 \: \land \: 3x \geq 2^x[/tex].

So far I've got:
[tex]x > 4 \: \rightarrow \: 2^x > 16 \quad[/tex] and [tex]\quad x > 4 \: \leftrightarrow \: 3x > 12[/tex].
But I am stuck on what to do next. Any suggestion would be appreciated.


Well let's try induction.

Let [tex]P_1 = x = 5 \Rightarrow 3x < 2^x[/tex]

[tex]15 < 2^5 = 32 \text{ is True}[/tex]

Now assume [tex]P_n \text{ is True, and show $P_{n+1}$ is as well}[/tex]

Assume for contradiction that [tex]3(x+1) > 2^{x+1}[/tex]

[tex]3(x+1) > 2^{x+1}\\

3x + 3 > 2 \cdot 2^x\\

3x - 2^x > 2^x - 3 \\

3x - 2^x < 0 \text{ because $P_n$ is True, so}\\
0 > 2^x - 3\\
\text{ and this is clearly False for $x > 1$, thus by contradiction}\\
\text{$3^{x+1} < 2^{x+1} \Rightarrow P_{n+1}$ is True, and $3x < 2^x,~\forall x > 4$}[/tex]

romsek
 
Posts: 12
Joined: Sun Feb 21, 2021 7:57 am
Reputation: 7

Re: Disproof of an existential proposition

Postby gio_osh » Thu Mar 25, 2021 12:49 am

Thanks for your answer.

On the last line, I think you meant:
$3(x+1) < 2^{x+1} \Rightarrow P_{n+1}$ is True, and $3x < 2^x,~\forall x > 4$

gio_osh
 
Posts: 2
Joined: Wed Mar 24, 2021 12:46 am
Reputation: 0

Re: Disproof of an existential proposition

Postby Guest » Mon Jul 12, 2021 6:58 pm

An English lesson- the verb is "disprove", not a "disproof".
Guest
 


Return to Inequalities



Who is online

Users browsing this forum: No registered users and 1 guest