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1990 IMO Problems/Problem 4 - Revision history
2024-03-28T23:23:43Z
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Proproblemsolver: /* Solution */
2022-09-30T15:01:09Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 15:01, 30 September 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l22" >Line 22:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Thus <math>f(xf(y)) = \frac{f(x)}{y}</math> QED</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Thus <math>f(xf(y)) = \frac{f(x)}{y}</math> QED</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>{{<del class="diffchange diffchange-inline">solution}</del>}</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==solution1==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">We show first that f(1) = 1. Taking x = y = 1, we have f(f(1)) = f(1). Hence f(1) = f(f(1)) = f(1 f(f(1)) ) = f(1)/f(1) = 1.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Next we show that f(xy) = f(x)f(y). For any y we have 1 = f(1) = f(1/f(y) f(y)) = f(1/f(y))/y, so if z = 1/f(y) then f(z) = y. Hence f(xy) = f(xf(z)) = f(x)/z = f(x) f(y).</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Finally, f(f(x)) = f(1 f(x)) = f(1)/x = 1/x.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">We are not required to find all functions, just one. So divide the primes into two infinite sets S = </ins>{<ins class="diffchange diffchange-inline">p1, p2, ... } and T= </ins>{<ins class="diffchange diffchange-inline">q1, q2, ... </ins>}<ins class="diffchange diffchange-inline">. Define f(pn) = qn, and f(qn) = 1/pn. We extend this definition to all rationals using f(xy) = f(x) f(y): f(pi1pi2...qj1qj2.../(pk1...qm1...)) = pm1...qi1.../(pj1...qk1...). It is now trivial to verify that f(x f(y)) = f(x)/y.  </ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See Also == {{IMO box|year=1990|num-b=3|num-a=5}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See Also == {{IMO box|year=1990|num-b=3|num-a=5}}</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Olympiad Algebra Problems]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Olympiad Algebra Problems]]</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Functional Equation Problems]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Functional Equation Problems]]</div></td></tr>
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Proproblemsolver
https://artofproblemsolving.com/wiki/index.php?title=1990_IMO_Problems/Problem_4&diff=175199&oldid=prev
Three14159265358979323846 at 03:43, 22 June 2022
2022-06-22T03:43:09Z
<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 03:43, 22 June 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9" >Line 9:</td>
<td colspan="2" class="diff-lineno">Line 9:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>f(1) = f(f(y) \times \frac{1}{f(y)}) = \frac{f(\frac{1}{f(y)})}{y} = 1</math>. This means that <math>f(x) = y</math>. Since <math>X,Y \in Q+</math> there will always be an <math>x</math> that suffices this and a corresponding <math>y</math>. Thus <math>f(ab)</math> with <math>b = f(c)</math> <math>\leftrightarrow</math> <math>c = \frac{1}{f(b)}</math> expands into <math>f(af(c)) = \frac{f(a)}{c} = f(a)f(b)</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>f(1) = f(f(y) \times \frac{1}{f(y)}) = \frac{f(\frac{1}{f(y)})}{y} = 1</math>. This means that <math>f(x) = y</math>. Since <math>X,Y \in Q+</math> there will always be an <math>x</math> that suffices this and a corresponding <math>y</math>. Thus <math>f(ab)</math> with <math>b = f(c)</math> <math>\leftrightarrow</math> <math>c = \frac{1}{f(b)}</math> expands into <math>f(af(c)) = \frac{f(a)}{c} = f(a)f(b)</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Utilizing <math>f(ab) = f(a)f(b)</math> we can see that if we break-down every ratio into it's prime factorization of the numerator and denominator. Our "4-chain" is simply <math>f(p) = q</math>, <math>f(q) = \frac{1}{p}</math>, <math>f(\frac{1}{p}) = \frac{1}{q}</math>, <math>f(\frac{1}{q}) = p</math> where <math>p</math> and <math>q</math> are unique primes. Thus our 2 sets <math>S_1</math> and <math>S_2</math> is simply 2 sets of equal size containing only primes. A good way to do this assuming we aren't given the sequence of all infinite primes (as this would require an explicit formula that does not yet exist) is to split the primes into 2 and 1 mod 6 being in one set with 3 and 5 mod 6 in the other. These sets have equal sizes from a strong form of Dirichlet's theorem on arithmetic progressions.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Utilizing <math>f(ab) = f(a)f(b)</math> we can see that if we break-down every ratio into it's prime factorization of the numerator and denominator. Our "4-chain" is simply <math>f(p) = q</math>, <math>f(q) = \frac{1}{p}</math>, <math>f(\frac{1}{p}) = \frac{1}{q}</math>, <math>f(\frac{1}{q}) = p</math> where <math>p</math> and <math>q</math> are unique primes. Thus our 2 sets <math>S_1</math> and <math>S_2</math> is simply 2 sets of equal size containing only primes <ins class="diffchange diffchange-inline">and their inverse</ins>. A good way to do this assuming we aren't given the sequence of all infinite primes (as this would require an explicit formula that does not yet exist) is to split the primes into 2 and 1 mod 6 being in one set with 3 and 5 mod 6 in the other. These sets have equal sizes from a strong form of Dirichlet's theorem on arithmetic progressions.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Thus we have created our two sets and to show that this actually works we will let <math>x = P_1^{E_{1x}} \times P_2^{E_{2x}} \times P_3^{E_{3x}} \times \dots</math> where <math>P_i</math> is a prime and <math>E_{ix}</math> are integers. Define <math>y</math> similarly. For the sake of simplicity if <math>i</math> is odd then <math>P_i \in S_1</math> and if <math>i</math> is even <math>P_i \in S_2</math>. Where our "4-chain" is <math>f(p_n) = p_{n+1}</math>, <math>f(p_{n+1}) = \frac{1}{p_n}</math>, <math>f(\frac{1}{p_n}) = \frac{1}{p_{n+1}}</math>, <math>f(\frac{1}{p_{n+1}}) = p_n</math> with <math>n</math> being odd. Then <math>f(y) = P_2^{E_{1y}} \times P_1^{-E_{2y}} \times P_4^{E_{3y}} \times P_3^{-E_{4y}}\dots</math> (remember <math>f(ab) = f(a)f(b)</math>).  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Thus we have created our two sets and to show that this actually works we will let <math>x = P_1^{E_{1x}} \times P_2^{E_{2x}} \times P_3^{E_{3x}} \times \dots</math> where <math>P_i</math> is a prime and <math>E_{ix}</math> are integers. Define <math>y</math> similarly. For the sake of simplicity if <math>i</math> is odd then <math>P_i \in S_1</math> and if <math>i</math> is even <math>P_i \in S_2</math>. Where our "4-chain" is <math>f(p_n) = p_{n+1}</math>, <math>f(p_{n+1}) = \frac{1}{p_n}</math>, <math>f(\frac{1}{p_n}) = \frac{1}{p_{n+1}}</math>, <math>f(\frac{1}{p_{n+1}}) = p_n</math> with <math>n</math> being odd. Then <math>f(y) = P_2^{E_{1y}} \times P_1^{-E_{2y}} \times P_4^{E_{3y}} \times P_3^{-E_{4y}}\dots</math> (remember <math>f(ab) = f(a)f(b)</math>).  </div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">\newline</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Then <math>xf(y) = P_1^{E_{1x} - E_{2y}} \times P_2^{E_{2x} + E_{1y}} \times P_3^{E_{3x} - E_{4y}} \times \dots</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Then <math>xf(y) = P_1^{E_{1x} - E_{2y}} \times P_2^{E_{2x} + E_{1y}} \times P_3^{E_{3x} - E_{4y}} \times \dots</math></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">\newline</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>f(xf(y)) = P_2^{E_{1x} - E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}} \dots</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>f(xf(y)) = P_2^{E_{1x} - E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}} \dots</math></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">\newline</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>f(x) = P_2^{E_{1x}} \times P_1^{-E_{2x}} \times P_4^{E_{3x}} \times P_3^{-E_{4x}}\dots</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>f(x) = P_2^{E_{1x}} \times P_1^{-E_{2x}} \times P_4^{E_{3x}} \times P_3^{-E_{4x}}\dots</math></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">\newline</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>\frac{f(x)}{y} = P_2^{E_{1x}- E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}}\dots</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><math>\frac{f(x)}{y} = P_2^{E_{1x}- E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}}\dots</math></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">\newline</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Thus <math>f(xf(y)) = \frac{f(x)}{y}</math> QED</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Thus <math>f(xf(y)) = \frac{f(x)}{y}</math> QED</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{solution}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{solution}}</div></td></tr>
</table>
Three14159265358979323846
https://artofproblemsolving.com/wiki/index.php?title=1990_IMO_Problems/Problem_4&diff=175198&oldid=prev
Three14159265358979323846 at 03:41, 22 June 2022
2022-06-22T03:41:17Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 03:41, 22 June 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l3" >Line 3:</td>
<td colspan="2" class="diff-lineno">Line 3:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution==</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">If we let <math>y = x = 1</math> this implies <math>f(f(1)) = f(1)</math>. Plugging <math>f(1) = y</math> in with this new fact. We get <math>f(f(f(1))) = \frac{f(1)}{f(1)} = 1</math>. Using <math>f(f(1)) = f(1)</math> again, we see that <math>f(f(f(1))) = f(f(1)) = f(1) = 1</math>. Now plugging in <math>x = 1</math> we get <math>f(f(y)) = \frac{1}{y}</math>. Such a function can not be continuous as if <math>f(x)</math> is increasing or decreasing on some observable interval, <math>f(f(x))</math> will be increasing on that interval but <math>\frac{1}{x}</math> is decreasing on all positive intervals. This indicates the function is discrete which means we can assign a "4-chain" of <math>f(a) = b</math>, <math>f(b) = \frac{1}{a}</math>, <math>f(\frac{1}{a}) = \frac{1}{b}</math>, and <math>f(\frac{1}{b}) = a</math>. It is obvious to see that any function where <math>f(f(x)) = \frac{1}{x}</math>. Must follow such a structure. The problem occurs that we do not know whether our current value of <math>x</math> is a <math>b</math> or an <math>a</math>. To make non-trivial progress on this we must split all rational numbers into two sets, <math>S_1</math> and <math>S_2</math>, such that if <math>E \in S_1</math> then <math>\frac{1}{E} \in S_1</math>. As long as <math>S_1</math> and <math>S_2</math> have the same size, so the bijection <math>f(\textrm{an element from }S_1)</math> = <math>(\textrm{an element from } S_2)</math> can hold, there will exist a piece-wise function (possibly with a greater than countable infinity number of pieces) that satisfies f(x). </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">To find one of these functions we can limit our searching by proving that <math>f(ab) = f(a)f(b)</math>. To do this simply set <math>x = \frac{1}{f(y)}</math> yielding:</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\newline</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>f(1) = f(f(y) \times \frac{1}{f(y)}) = \frac{f(\frac{1}{f(y)})}{y} = 1</math>. This means that <math>f(x) = y</math>. Since <math>X,Y \in Q+</math> there will always be an <math>x</math> that suffices this and a corresponding <math>y</math>. Thus <math>f(ab)</math> with <math>b = f(c)</math> <math>\leftrightarrow</math> <math>c = \frac{1}{f(b)}</math> expands into <math>f(af(c)) = \frac{f(a)}{c} = f(a)f(b)</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Utilizing <math>f(ab) = f(a)f(b)</math> we can see that if we break-down every ratio into it's prime factorization of the numerator and denominator. Our "4-chain" is simply <math>f(p) = q</math>, <math>f(q) = \frac{1}{p}</math>, <math>f(\frac{1}{p}) = \frac{1}{q}</math>, <math>f(\frac{1}{q}) = p</math> where <math>p</math> and <math>q</math> are unique primes. Thus our 2 sets <math>S_1</math> and <math>S_2</math> is simply 2 sets of equal size containing only primes. A good way to do this assuming we aren't given the sequence of all infinite primes (as this would require an explicit formula that does not yet exist) is to split the primes into 2 and 1 mod 6 being in one set with 3 and 5 mod 6 in the other. These sets have equal sizes from a strong form of Dirichlet's theorem on arithmetic progressions.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Thus we have created our two sets and to show that this actually works we will let <math>x = P_1^{E_{1x}} \times P_2^{E_{2x}} \times P_3^{E_{3x}} \times \dots</math> where <math>P_i</math> is a prime and <math>E_{ix}</math> are integers. Define <math>y</math> similarly. For the sake of simplicity if <math>i</math> is odd then <math>P_i \in S_1</math> and if <math>i</math> is even <math>P_i \in S_2</math>. Where our "4-chain" is <math>f(p_n) = p_{n+1}</math>, <math>f(p_{n+1}) = \frac{1}{p_n}</math>, <math>f(\frac{1}{p_n}) = \frac{1}{p_{n+1}}</math>, <math>f(\frac{1}{p_{n+1}}) = p_n</math> with <math>n</math> being odd. Then <math>f(y) = P_2^{E_{1y}} \times P_1^{-E_{2y}} \times P_4^{E_{3y}} \times P_3^{-E_{4y}}\dots</math> (remember <math>f(ab) = f(a)f(b)</math>). </ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\newline</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Then <math>xf(y) = P_1^{E_{1x} - E_{2y}} \times P_2^{E_{2x} + E_{1y}} \times P_3^{E_{3x} - E_{4y}} \times \dots</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\newline</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>f(xf(y)) = P_2^{E_{1x} - E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}} \dots</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\newline</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>f(x) = P_2^{E_{1x}} \times P_1^{-E_{2x}} \times P_4^{E_{3x}} \times P_3^{-E_{4x}}\dots</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\newline</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"><math>\frac{f(x)}{y} = P_2^{E_{1x}- E_{2y}} \times P_1^{-E_{2x} - E_{1y}} \times P_4^{E_{3x} - E_{4y}} \times P_3^{-E_{4x} - E_{3y}}\dots</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">\newline</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Thus <math>f(xf(y)) = \frac{f(x)}{y}</math> QED</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{solution}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{solution}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
</table>
Three14159265358979323846
https://artofproblemsolving.com/wiki/index.php?title=1990_IMO_Problems/Problem_4&diff=144075&oldid=prev
Hamstpan38825 at 17:46, 30 January 2021
2021-01-30T17:46:55Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 17:46, 30 January 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">4. </del>Let <math>\mathbb{Q^+}</math> be the set of positive rational numbers. Construct a function <math>f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}</math> such that <math>f(xf(y)) = \frac{f(x)}{y}</math> for all <math>x, y\in{Q^+}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==Problem==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Let <math>\mathbb{Q^+}</math> be the set of positive rational numbers. Construct a function <math>f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}</math> such that <math>f(xf(y)) = \frac{f(x)}{y}</math> for all <math>x, y\in{Q^+}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">==Solution==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">{{solution}}</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">== See Also == {{IMO box|year=1990|num-b=3|num-a=5}}</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Olympiad Algebra Problems]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Olympiad Algebra Problems]]</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Functional Equation Problems]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Functional Equation Problems]]</div></td></tr>
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Hamstpan38825
https://artofproblemsolving.com/wiki/index.php?title=1990_IMO_Problems/Problem_4&diff=79568&oldid=prev
1=2 at 12:44, 19 July 2016
2016-07-19T12:44:07Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 12:44, 19 July 2016</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>4. Let <math>\mathbb{Q^+}</math> be the set of positive rational numbers. Construct a function <math>f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}</math> such that <math>f(xf(y)) = \frac{f(x)}{y}</math> for all <math>x, y\in{Q^+}</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>4. Let <math>\mathbb{Q^+}</math> be the set of positive rational numbers. Construct a function <math>f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}</math> such that <math>f(xf(y)) = \frac{f(x)}{y}</math> for all <math>x, y\in{Q^+}</math>.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">[[Category:Olympiad Algebra Problems]]</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">[[Category:Functional Equation Problems]]</ins></div></td></tr>
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1=2
https://artofproblemsolving.com/wiki/index.php?title=1990_IMO_Problems/Problem_4&diff=79178&oldid=prev
Ani2000 at 09:51, 5 July 2016
2016-07-05T09:51:00Z
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 09:51, 5 July 2016</td>
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Ani2000
https://artofproblemsolving.com/wiki/index.php?title=1990_IMO_Problems/Problem_4&diff=79177&oldid=prev
Ani2000: Created page with "4. Let <math>\mathbb{Q^+}</math> be the set of positive rational numbers. Construct a function <math>f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}</math> such that <math>f(xf(y)) = \..."
2016-07-05T09:50:31Z
<p>Created page with "4. Let <math>\mathbb{Q^+}</math> be the set of positive rational numbers. Construct a function <math>f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}</math> such that <math>f(xf(y)) = \..."</p>
<p><b>New page</b></p><div>4. Let <math>\mathbb{Q^+}</math> be the set of positive rational numbers. Construct a function <math>f :\mathbb{Q^+}\rightarrow\mathbb{Q^+}</math> such that <math>f(xf(y)) = \frac{f(x)}{y}</math> for all $x, y\in{Q^+}.</div>
Ani2000