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]