This is the second part in a 3-part blog post in which we prove that is transcendental.

* Three-step proof that is transcendental*Step 1

**Step 2**

Step 3

Recall that in step 1 we proved the following lemma.

**Lemma 1.** Suppose is a root of the polynomial . Let be a polynomial and . Then there exist such that , where .

** Step 2.**

Now we are ready pick a specific polynomial for Lemma 1. Since there are infinitely many primes, we can find a prime number greater than both and . Let

Our goal for this step is to show that is a nonzero integer. In fact we will show is that it is an integer not divisible by , which implies that it is nonzero.

We will accomplish this by proving that for , (and hence ) is an integer divisible by , but that is an integer that is not divisible by .

First we state but do not prove the following lemma. (This is a homework problem in Herstein.)

**Lemma 2.** If is a polynomial with integer coefficients and , then for , is a polynomial with integer coefficients, each of which is divisible by .

Applying Lemma 2 to our function we see that when , is a polynomial with integer coefficients, each of which is divisible by . In particular, when and is any integer, is an integer that is divisible by .

Now let us restrict our attention to . From the definition of we see that is a root of with multiplicity . The following lemma tells us that . (We leave the proof of this lemma as an exercise for the reader.)

**Lemma 3.** Suppose is a root of a polynomial with multiplicity . Then for .

So,

We know from above that each of the terms is an integer divisible by . Thus is an integer divisible by .

Now let us consider . By the definition of , is a root with multiplicity . By Lemma 3, .

Thus,

From the result above, we know that are all divisible by . What about ?

If we multiply out the terms in , we obtain

So,

and hence . Since , does not divide . Therefore does not divide . Moreover, , so does not divide . We conclude that does not divide .

Finally this implies that is an integer not divisible by . In particular, it is a nonzero integer.

We are almost finished with the proof. In the next step we will show that , thus it cannot be a nonzero integer and we will obtain our sought-after contradiction.

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The transcendence of e « Division by Zeroon September 28, 2010at 3:30 pm

Thanks for posting this, it’s very helpful. I have one question though, I know I’m being stupid but I can’t seem to prove lemma 2. I’d really appreciate any help on this. Thank you.

By:

David Smithon September 5, 2011at 4:53 pm

For simplicity, suppose . Then . Rewritten, the coefficient is . Seen in this way, it is clearly an integer divisible by .

By:

Dave Richesonon September 5, 2011at 5:28 pm

Thanks a lot for your help, I really should have been able to work that out myself!

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David Smithon September 6, 2011at 4:29 pm

Sorry to bother you again, but I’ve been trying to prove that $e^{m/n}$ is transcendental, where $m>0$ and $n$ are integers (it’s an exercise in Herstein on p.178), but I’ve had no luck. Any help would be much appreciated. Thanks.

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David Smithon December 9, 2011at 11:25 am