+0  
 
0
238
4
avatar

I need help determining which integer is closest to $(1+\sqrt 2)^8$

Thanks!

 May 7, 2022
 #1
avatar
-1

The integer closest to (1 + sqrt(2))^8 is 1648.

 May 7, 2022
 #2
avatar
0

That was incorrect. Are you sure?

Guest May 7, 2022
 #3
avatar
0

The answer was 1154.

 May 7, 2022
 #4
avatar+9459 
0

Observation: \((1 + \sqrt 2)^8 + (\sqrt 2-1)^8\) is an integer, and \(0 < (\sqrt 2- 1)^8 < \dfrac12\). That integer is the required integer.

 

Using binomial theorem,

\(\quad(1 + \sqrt 2)^8 + (\sqrt 2-1)^8\\ =\displaystyle\sum_{k = 0}^8 \binom{8}k ((\sqrt 2)^k + (-\sqrt 2)^k)\\ =\displaystyle\binom{8}0 (2) + \binom82 (2(\sqrt 2)^2) + \binom84 (2(\sqrt 2)^4) + \binom86 (2(\sqrt 2)^6) + \binom88 (2(\sqrt 2)^8)\\ = 1154\)

 

Note that \(\sqrt 2 - 1 \approx 0.414\), so \(0 < (\sqrt 2 - 1)^8 < \sqrt 2 - 1 < \dfrac12\). That means \((1 + \sqrt 2)^8\) plus a real number way smaller than 1/2 gives 1154.

 

Therefore, the nearest integer to \((1 + \sqrt 2)^8\) is 1154.

 

Remarks: The error is \(\left|(1 + \sqrt2)^8 - 1154\right| = (\sqrt 2 - 1)^8 \approx 8.665517772 \times 10^{-4}\), which is really really small.

 May 9, 2022

9 Online Users

avatar