Subject: proof of inequality by mathematical induction Name: Carol Who are you: Student S(n) = 2^n > 10n+7 and n>=10 Basis step is true: S(10) is true I assume S(k) is true: S(k) = 2^k > 10k+7 Have to show s(k+1) = 2^k+1 > 10(k+1) + 7 is true. This is where I get stuck. Hi Carol, Good work so far. If S(k) > 10k+7, as you've assumed, then 2 S(k) > 2(10k + 7) but since 2(10k + 7) = 10k + 10 + 7 + 10k - 3 = 10(k+1) + 7 + 10k - 3 then 2 S(k) > 10(k+1) + 7 + 10k - 3 and since k ≥ 10, then 10k - 3 > 0, so 10(k+1) + 7 + 10k - 3 > 10(k+1) + 7 So if a > b > c then a > c: 2 S(k) > 10(k+1) + 7 and since S(k+1) = 2^(k+1) = 2 2k = 2 S(k), we have proven: S(k+1) > 10(k+1) + 7. Stephen La Rocque.> In January 2019 Gary wrote : This problem was done by Stephen La Rocque. It is 2^n > 10n + 7. I am confused about the line that reads: but since 2(10k +7) = 10k + 10 + 7 + 10k -3 = 10(k + 1) + 7 + 10 -3. This seems to be very confusing. Can Stephen or anyone on your staff shed some light on this for me? How did you arrive at these quantities? Hi Gary, Stephen responded to this question in 2006 when he was a university student. He is no longer here but I'll try to answer your question. It is difficult to know what he was thinking. The induction step is to assume that $2^k > 10k + 7$ and from this prove that $2^{k+1} > 10(k+1) + 7.$ $2^{k+1} = 2 \times 2^k$ so stephen calculated $2 \times 2^k$ and got $2^{k+1} = 2 \times 2^k = 2(10k + 7) = 10 k + 7 + 10k + 7.$ He then compared the right side to the right side of wat he wanted $2^{k+1} > 10(k+1) + 7$ and realized that if he wrote one on the sevens in 10 k + 7 + 10k + 7 as 10 - 3 he would have $2^{k+1} = 10 k + 7 + 10k + 7 = 10k + 10 - 3 + 10k + 7 = 10(k+1) + 7 + 10k - 3.$ All that was left to do to complet the induction step was to show that $10k - 3 > 0.$ I hope this helps,Harley