New PDF release: A Course on Number Theory [Lecture notes]

By Peter J. Cameron

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So qn−1 y − pn−1 and qn y − pn have opposite signs. So u(qn−1 y − pn−1 ) and v(qn y − pn ) have the same sign. It follows that the absolute value of their sum is the sum of their absolute values: |qy − p| = = = = > |(qn−1 u + qn v)y − (pn−1 u + pn v)| |u(qn−1 y − pn−1 ) + v(qn y − pn )| |u(qn−1 y − pn−1 )| + |v(qn y − pn )| |u| · |qn−1 y − pn−1 | + |v| · |qn y − pn | |qn−1 y − pn−1 |, where in the last step we use the fact that |u| ≥ 1 and |v| · |qn y − pn | > 0. This completes the proof. 1 Find the first six terms in the continued fraction for the number e, the root of natural logarithms.

5 Suppose that √ n = [a0 ; a1 , . . , ak , 2a0 ] √ where k is odd, say, k = 2m + 1. Write ym+1 = (Pm+1 + n)/Qm+1 . Then 2 n = Pm+1 + Q2m+1 . Proof In this case, y = [a0 ; a1 , . . , am , am , . . , a1 , 2a0 ]. We have ym+1 = [am+1 , . . , a2m , 2a0 , a1 , . . , am ], and −1/ym+1 = [am ; . . , a1 , 2a0 , a2m , . . , am+1 ]. But by assumption, these two continued fractions are identical. So ym+1 ym+1 = −1. √ √ Now ym+1 = (Pm+1 + n)/Qm+1 and ym+1 = (Pm+1 − n)/Qm+1 . So we have 2 Pm+1 −n = −1, 2 Qm+1 2 so n = Pm+1 + Q2m+1 , as required.

An , y] = ypn + pn−1 , yqn + qn−1 so qn y2 + (qn−1 − pn )y + pn−1 = 0. Let z = [an ; an−1 , . . , a0 ]. Then z[an , . . , a0 ] + [an , . . , a1 ] z[an−1 , . . , a0 ] + [an−1 , . . , a1 ] zpn + qn = , zpn−1 + qn−1 z = [an ; an−1 , . . , a0 , z] = where we use the fact that [a0 , . . , an ] = [an , . . , a0 ]. So pn−1 z2 + (qn−1 − pn )z − qn = 0. In other words, qn (−1/z)2 + (qn−1 − pn )(−1/z) + pn−1 = 0. Thus, −1/z satisfies the same quadratic equation as y. But z > 1, so −1 < −1/z < 0, whereas y > 1; so −1/z is the other root y of the quadratic equation.

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