Use foreword reasoning to show that if x is a nonzero real number then x2 + 1/x2
ID: 2945012 • Letter: U
Question
Use foreword reasoning to show that if x is a nonzero real number then x2 + 1/x2 2.[Hint: Satar with the in equality(x-1/x2) 0 which holds for all nonzero real numbers x.] The harmonic mean of two real number x and y 2xy/(x + y). By computing the harmonic ac and late a conjecture about their relative sizes and conjecture. The quadratic men of two real numbers x and y equal By computing the arithmetic and quadratic means of different pair of positive real numbers formulate a conjecture about their relative sizes and prove your conjecture. Write the numbers 1, 2...,2n on a blackboard. Where a is an odd integer. Pick any of the number, f and k, write |j-k|on board and erase j and k. Continue this process until only one integer is written on the board. Prove that this integer must be odd. Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any two unequal bits you insert a 1 to produce nine new bits. Then you erase the nine original bits. Show that when you iterate this procedure, you can never get nine zero.[Hint: Work backward, assuming that you did end up with nine zero.] Formulate a conjecture about the decimal the digits that appear as the final decimal digit of the fourth power of an integer. Prove your conjecture using a proof by cases.Explanation / Answer
Let Sk represent the sum of the numbers on the board after k steps.
At first, S0=1+2+...+2n = (2n)(2n+1)/2 = n(2n+1). The number n is, by assumption, an odd number, and 2n+1 is always odd, so S0 is odd.
At each step, we replace to integers, i and j, with |i-j|. Without loss of generality, let ij, and so |i-j| = i-j.
Then Sk+1 = Sk - (what we are taking off the board) + (what we are adding to the board)
= Sk - (i+j) + (i-j)
= Sk - i - j + i - j
= Sk -2j
In other words, Sk and Sk+1 differ by 2j, an even number. This means they have the same parity (both are even or both are odd).
We have shown that S0 is odd. Therefore Sk must be odd until the very end, which will be after 2n-1 terms.
However, at the end, all we are left with is a single number. In other words, S2n-1 is just going to equal the last number on the board. However, we have shown that S2n-1 must be odd! Therefore the last number is odd.
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