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Correction: Problem Solving: Tips For TeachersSource: The Arithmetic Teacher, Vol. 33, No. 4 (December 1985), p. 48Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/80000190 .Accessed: 12/06/2014 12:44Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp .JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org. .National Council of Teachers of Mathematics is collaborating with JSTOR to digitize, preserve and extendaccess to The Arithmetic Teacher.http://www.jstor.org This content downloaded from 91.229.229.44 on Thu, 12 Jun 2014 12:44:30 PMAll use subject to JSTOR Terms and Conditionshttp://www.jstor.org/action/showPublisher?publisherCode=nctmhttp://www.jstor.org/stable/80000190?origin=JSTOR-pdfhttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jspReader*1 DICltOQIK! [Continued from page 26) The crossword answer space shown here is not complete. Only some of the blackened squares are provided. Using the information in the previous paragraph, locate the rest of the squares. Answer: William D. Ja m ski Indiana University outheast New Albany, IN 47150 Insights into density In the discussion of the density property of the rational numbers with preservice elemen- tary teachers, I usually ask, "Can you find a rational number midway between two rational numbers?" We then discuss the averaging method, along with the fact that it is always possible to find a rational number between any two rational numbers. In my last discussion, a student spoke up, "I've found another way to get a rational number between two rational numbers. I just add the numerators to find the new numerator and add the denominators to get the new de- nominator." The question from the class was, "Does it always work?" Examples were test- ed by using the fact that if alb < eld where a, b, c, and d are integers and b and d are posi tive, then ad < be. We ended up writing a proof as follows: If a, b, and d are integers and b and d are positive and alb < eld, then ad < be. If we assume that a a + ~b< b + then we know that a (b + a) < b(a + c), or ab + ad < ab + be. Subtracting ab from each side of the inequal- ity results in ad < be, which was given. Similarly, if we assume that a + d + d