Transcription of SOLUTIONS - Springer
{{id}} {{{paragraph}}}
Example: air traffic controller
SOLUTIONS - Springer
SOLUTIONSM ethods of the contrary, namely that 2+ 3+ 5=r, whereris a rational the equality 2+ 3=r 5 to obtain 5+2 6=r2+5 2r 5. It followsthat 2 6+2r 5 is itself rational. Squaring again, we find that 24+20r2+8r 30is rational, and hence 30 is rational, too. Pythagoras method for proving that 2isirrational can now be applied to show that this is not true. Write 30=mnin lowestterms; then transform this intom2=30n2. It follows thatmis divisible by 2 and because2(m2)2=15n2it follows thatnis divisible by 2 as well. So the fraction was not in lowestterms, a contradiction. We conclude that the initial assumption was false, and therefore 2+ 3+ 5 is that such numbers do exist, and let us look at their prime factorizations.
(German Mathematical Olympiad, 1985) 5. Arguing by contradiction, let us assume that the area of the overlap of any two sur-faces is less than 1 9. In this case, if S 1,S 2,...,S n denote the nine surfaces, then the area of S 1 ∪S 2 is greater than 1 +8 9, the area of S 1 ∪S 2 ∪S 3 is greater than 1 +8 9 +7 9,..., and the area of S 1 ∪S ...
Loading..
Tags:
Information
Domain:
Source:
Link to this page:
Please notify us if you found a problem with this document: