#### ThunderBase

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- Thread starter ThunderBase
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Do you agree that: If a<b then 2a+2 < 2b+2 ?

That's pretty much what they've done if you replace "a" with 2^k+2k and "b" with k!.

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We have 2*k! + 2, and want to get to (k+1)! (that is, the former has to be less than the latter). How can we link them? Well, expressing the latter in terms of k!, it is (k + 1)k!, so that's the expression we need at this point; we just have to justify the claim that 2*k! + 2 < (k + 1)k!.

So the hard part is actually the "reason" part (and filling the gap as we read, to see that that is actually true). Can you see why k >= 5 implies 2*k! + 2 < (k + 1)k! ? Give it a try. (I personally wish they had written at least one more step in there.)

Like others, I find this proof excessively concise.Hi All,

I'm having trouble understanding how to solve this question. I don't understand how they got those steps that I highlighted below.

Any help is appreciated.

Thanks!

View attachment 11922

We know that the proposition to be proved is

\(\displaystyle \text {Let } k \text { be ANY integer } \ge 5 \text { such that } 2^k + 2 < k!.\)

\(\displaystyle 2^{(k+1)} + 2(k + 1) = 2^k * 2^1 + 2k + 2 = 2 * 2^k + 2k + 2.\)

\(\displaystyle \therefore 2^{(k+1)} + 2(k + 1) = (2 * 2^k + 2k + 2) < (2 * 2^k + 2k + 2) + 2k \ \because \ 5 \le k.\)

\(\displaystyle \therefore 2^{(k+1)} + 2(k + 1) < 2 * 2^k + 4k + 4k + 2 = 2(2^k + 2k) + 2.\)

\(\displaystyle \text {But } 2^k + 2 < k! \text { by first line's definition of } k.\)

\(\displaystyle \therefore 2^{(k+1)} + 2(k + 1) < 2k! + 2.\)

\(\displaystyle \text {And } k! > k > 2 \ \because k \ge 5 \text { by definition.}\)

\(\displaystyle 2^{(k+1)} + 2(k + 1) < 2k! + 2 < k * k! + k!.\)

\(\displaystyle \text {THUS, } 2^{(k+1)} + 2(k + 1) < 2k! + 2 < k * k! + k! = k!(k + 1) = (k + 1)! \text { Q.E.D.}\)

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Thank you all. Makes perfect sense now!