Algebra 2 unit 5
... We have looked at (and solved) quadratic equations that either have 2 solutions (and cross the x-axis twice) or 1 solution (cross the x-axis once). Now, what would the quadratic formula look like for a graph that has no real solutions, meaning that the graph never crosses the x-axis and the solution ...
... We have looked at (and solved) quadratic equations that either have 2 solutions (and cross the x-axis twice) or 1 solution (cross the x-axis once). Now, what would the quadratic formula look like for a graph that has no real solutions, meaning that the graph never crosses the x-axis and the solution ...
Solutions 7
... Proof. Clearly 1 is a perfect square. Suppose the claim works for integers up to and including k. Then if we write k + 1 = ab, the IH tells us that a = m2 and b = l2 for some integers m, l. Thus, k + 1 = m2 l2 = (ml)2 and k + 1 is a perfect square. Thus, by induction we conclude that all integers ar ...
... Proof. Clearly 1 is a perfect square. Suppose the claim works for integers up to and including k. Then if we write k + 1 = ab, the IH tells us that a = m2 and b = l2 for some integers m, l. Thus, k + 1 = m2 l2 = (ml)2 and k + 1 is a perfect square. Thus, by induction we conclude that all integers ar ...
A Nonlinear Expression for Fibonacci Numbers and Its Consequences
... This is precisely equivalent to the well-known theorem: Fm1 ···ms −1 ≡ 0 (mod Fm1 −1 · · · Fms −1 ) ...
... This is precisely equivalent to the well-known theorem: Fm1 ···ms −1 ≡ 0 (mod Fm1 −1 · · · Fms −1 ) ...