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
Like others, I find this proof excessively concise.
We know that the proposition to be proved is
TRUE for
AT LEAST one integer greater than or equal to 5, namely 5 itself. Therefore it is perfectly reasonable to say
[MATH]\text {Let } k \text { be ANY integer } \ge 5 \text { such that } 2^k + 2 < k!.[/MATH]
WE KNOW that there is such an integer. Now we want to show that the proposition is true for k + 1, and obviously
[MATH]2^{(k+1)} + 2(k + 1) = 2^k * 2^1 + 2k + 2 = 2 * 2^k + 2k + 2.[/MATH]
[MATH]\therefore 2^{(k+1)} + 2(k + 1) = (2 * 2^k + 2k + 2) < (2 * 2^k + 2k + 2) + 2k \ \because \ 5 \le k.[/MATH]
[MATH]\therefore 2^{(k+1)} + 2(k + 1) < 2 * 2^k + 4k + 4k + 2 = 2(2^k + 2k) + 2.[/MATH]
[MATH]\text {But } 2^k + 2 < k! \text { by first line's definition of } k.[/MATH]
[MATH]\therefore 2^{(k+1)} + 2(k + 1) < 2k! + 2.[/MATH]
[MATH]\text {And } k! > k > 2 \ \because k \ge 5 \text { by definition.}[/MATH]
[MATH]2^{(k+1)} + 2(k + 1) < 2k! + 2 < k * k! + k!.[/MATH]
[MATH]\text {THUS, } 2^{(k+1)} + 2(k + 1) < 2k! + 2 < k * k! + k! = k!(k + 1) = (k + 1)! \text { Q.E.D.}[/MATH]
