Timeline for answer to Exercise 1(d) from Courant by André Nicolas
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Aug 29, 2013 at 15:08 | vote | accept | CommunityBot | ||
| Aug 15, 2013 at 19:40 | comment | added | André Nicolas | @confusedmike: Solve many problems! The more you do, the easier things get, ideas can be recycled. | |
| Aug 15, 2013 at 19:35 | comment | added | user90438 | Wow, that's so awesome. Now how did you come upon that strategy? Intuition? Experimentation? I'm trying to get better at problem-solving in general, so I'd be interested to pick your brain here...thanks so much! | |
| Aug 15, 2013 at 19:29 | history | edited | André Nicolas | CC BY-SA 3.0 |
added 635 characters in body
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| Aug 15, 2013 at 19:19 | comment | added | André Nicolas | Or else if you want to bypass the suggested approach, go through the first two steps, then solve to get $\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}$. The number $x$ cannot be rational because if it were, from the last expression $\sqrt{2}$ would also be rational. However, we know it is not. | |
| Aug 15, 2013 at 19:13 | comment | added | André Nicolas | Not the best or intended way to do it. Expand $(x-\sqrt{2})^3=2$. We get $x^3 -3\sqrt{2}x^2+6x-2\sqrt{2}=2$. Rewrite as $x^3+6x-2=(3x^2+2)\sqrt{2}$. Now square both sides, and simplify. You will get a degree $6$ equation with integer coefficients. | |
| Aug 15, 2013 at 19:07 | comment | added | user90438 | Oh I see...excellent! Now, it seems to me that one has to take $x - \sqrt{2}$ up to a power of 6 in order to come up with the target 6-th order polynomial. In expanded form, I get $$x^6 + 6\sqrt{2}x^5 + 30x^4 + 40\sqrt{2}x^3 + 60x^2 + 24\sqrt{2}x + 4$$ and I think your logic works from here? | |
| Aug 15, 2013 at 5:48 | history | answered | André Nicolas | CC BY-SA 3.0 |