How to show properly that $\left( 1-\sqrt {2}\right) ^{3000} <\left( \frac {1}{10}\right) ^{100}$?

Question

I would like to show, $\left( 1-\sqrt {2}\right) ^{3000} <\left( \dfrac {1}{10}\right) ^{100}$

With a calculator I can do $\left( 1-\sqrt {2}\right) ^{2x} =\left( \dfrac {1}{10}\right) ^{100} $

$x = \dfrac {-100\log 10}{\log \left( 1-2\sqrt {2}+2\right) } ≈ 131 $

so I can see it is comfortably true, but how do I quickly and rigorously show this without a calculator?

Answer 1

Clearly,$$\left(1-\sqrt2\right)^{3000}<\left(\frac1{10}\right)^{100}\iff\left(\sqrt2-1\right)^{\frac{3000}{100}}<\frac1{10}\iff\left(\sqrt2-1\right)^{30}<\frac1{10}.$$But $\sqrt2-1<\frac12$ and therefore all you need to prove is that$$\left(\frac12\right)^{30}<\frac1{10},$$which is equivalent to $2^{30}>10$.

Answer 2

Alternatively, note that $(1-\sqrt{2})(1+\sqrt{2})=-1$, so $$(1-\sqrt{2})^{3000}<\left(\frac{1}{10}\right)^{100} \qquad\Leftrightarrow\qquad (1+\sqrt{2})^{3000}>10^{100}.$$ Of course this is equivalent to $(1+\sqrt{2})^{30}>10$, which is obvious as $1+\sqrt{2}>2$.

Answer 3

$(1-√2)= (1-√2) \dfrac{1+√2}{1+√2}= \dfrac{-1}{1+√2};$

$((1-√2)^{30})^{100} = $

$\dfrac{1}{(1+√2)^{30})^{100}} \lt$

$\dfrac {1}{(2^{30})^{100}} \lt \dfrac{1}{10^{100}}.$

Used:

$1+√2>2$, and

$2^{30} \gt 10.$