0

Could somebody help me with this induction?

$1 \cdot 2 \cdot 3 \cdot 4 ... n>2^n , n \ge 4$ , n is an integer n.

I tried to solve it. Sorry if my handwriting is not clear.

Sorry, I meant 4>1

JMoravitz
  • 79,518
Flori
  • 11
  • 2
  • You have the right idea however the way you formatted your proof is poor. It is a common mistake to see someone try to format a proof as you had by starting with what they hope to be true, manipulating both sides of the equation to get to the next line, manipulating both sides of that to get to the next and so on until arriving at a tautology. So common in fact, that I have encountered teachers who teach this as a method. It is improper and in some circumstances outright invalid in the case that the manipulations are not reversible. – JMoravitz Feb 04 '21 at 20:06
  • 1
    It is far better to format the induction step here as $1\cdot 2\cdot 3\cdots n\cdot (n+1) > \dots \geq \dots \geq \dots \geq 2^{n+1}$ all in a single line. See more about formatting an induction proof here. – JMoravitz Feb 04 '21 at 20:07
  • Another nitpick., you need to write $k+1$ if it appears in a product enclosed in parentheses. If you write $k\cdot k+1$ this looks more like $(k\cdot k)+1$ than it does $k\cdot (k+1)$. – JMoravitz Feb 04 '21 at 20:09
  • @JMoravitz Agreed, which is why I just deleted my answer. Actually, once I saw that the OP wasn't using mathJax and that the image was hard to read, I ignored the image. However, your point is well taken. – user2661923 Feb 04 '21 at 20:15

1 Answers1

1

The image of your solution suggests that you were in the right track, but as you reached the 3rd step, you made a mistake.

For proof by induction, we need to prove the following things:

  1. Check if $P(n_1)$ is true.
  2. Suppose that $P(k)$ is true.
  3. Prove that $P(k+1)$ is true.

Now, $$P(4) = 1*2*3*4 = 24 > 2^4$$ Thus, $P(4)$ is true.

Let,$$P(k) = 1*2*3...*k > 2^k$$

Now,$$P(k+1) = 1*2*3...*k*(k+1)$$ Since, $k≥4$,

therefore, $k+1>4$.

Now, multiply $k+1$ in $P(k)$.

Thus, $$1*2*3...*k*(k+1) > 2^k *(k+1)$$

Now, using the fact that if,

$$a>b$$

Then, $$K*a>K*b [for K>0]$$

We see that,

$$1*2*3...*k*(k+1) > 2^k*4$$

$$P(k+1) > 2^{k+2}$$ Thus,$$P(k+1) > 2^{k+1}$$ Hence, Proved!

Nandeesh Bhatrai
  • 616