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Chapter 1: Problem 115
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We call a sequence \(\left(x_n\right)_{n \geq 1}\) a superinteger if (i) each \(x_n\) is a nonnegative integer less than \(10^n\) and (ii) the last \(n\) digits of \(x_{n+1}\) form \(x_n\). One example of such a sequence is \(1,21,021,1021,21021,021021, \ldots\), which we abbreviate by ...21021. Note that the digit 0 is allowed (as in the example) and that (unlike in the example) there may not be a pattern to the digits. The ordinary positive integers are just those superintegers with only finitely many nonzero digits. We can do arithmetic with superintegers; for instance, if \(x\) is the superinteger above, then the product \(x y\) of \(x\) with the superinteger \(y=\ldots 66666\) is found as follows: \(1 \times 6=6\) : the last digit of \(x y\) is 6 . \(21 \times 66=1386\) : the last two digits of \(x y\) are 86 . \(021 \times 666=13986\) : the last three digits of \(x y\) are 986 . \(1021 \times 6666=6805986\) : the last four digits of \(x y\) are 5986, etc. Is it possible for two nonzero superintegers to have product \(0=\ldots 00000\) ?
Let \(R\) be a commutative ring with at least one, but only finitely many, (nonzero) zero divisors. Prove that \(R\) is finite.
For three points \(P, Q\), and \(R\) in \(\mathbb{R}^3\) (or, more generally, in \(\mathbb{R}^n\) ) we say that \(R\) is between \(P\) and \(Q\) if \(R\) is on the line segment connecting \(P\) and \(Q\) ( \(R=P\) and \(R=Q\) are allowed). A subset \(A\) of \(\mathbb{R}^3\) is called convex if for any two points \(P\) and \(Q\) in \(A\), every point \(R\) which is between \(P\) and \(Q\) is also in \(A\). For instance, an ellipsoid is convex, a banana is not. Now for the problem: Suppose \(A\) and \(B\) are convex subsets of \(\mathbb{R}^3\). Let \(C\) be the set of all points \(R\) for which there are points \(P\) in \(A\) and \(Q\) in \(B\) such that \(R\) lies between \(P\) and \(Q\). Does \(C\) have to be convex?
a. If a rational function (a quotient of two real polynomials) takes on rational values for infinitely many rational numbers, prove that it may be expressed as the quotient of two polynomials with rational coefficients. b. If a rational function takes on integral values for infinitely many integers, prove that it must be a polynomial with rational coefficients.
In general, composition of functions is not commutative. For example, for the functions \(f\) and \(g\) given by \(f(x)=x+1, g(x)=2 x\), we have \(f(g(x))=2 x+1\) and \(g(f(x))=2 x+2\). Now suppose that we have three functions \(f, g, h\). Then there are six possible compositions of the three, given by \(f(g(h(x))), g(h(f(x))), \ldots\). Give an example of three continuous functions that are defined for all real \(x\) and for which exactly five of the six compositions are the same. (Reprinted with the permission of the Canadian Mathematical Society, this problem was originally published in the Mathematical Mayhem section of Crux Mathematicorum with Mathematical Mayhem, vol. 25,1999, p. 293, problem C87.)
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