Pie?

Number Theory Level 4

$\pi$ and $e$ are both transcendental numbers. Transcendental numbers are numbers which can't be roots of a non-zero polynomial equation with rational coefficients. Keeping that in mind, read the following statements.

$[1]$ . It is possible for $\pi+e$ to be a rational number in base $10$ .

$[2]$ . At least one of $\pi+e$ and $\pi \times e$ is irrational in base $10$ .

$[3]$ . $\pi$ is an irrational number in base $10$ .

Which of these statements are correct?

Note : This problem is a part of the set "I Don't Have a Good Name For This Yet". See the rest of the problems here . And when I say I don't have a good name for this yet, I mean it. If you like problems like these and have a cool name for this set, feel free to comment here .

[1]

[2]

and

[3]

[3]

[2]

and

[3]

[1]

1 solution

Mursalin Habib
Apr 8, 2014

Let's consider the statements one by one.

$[1]$ Is it possible for $\pi+e$ to be a rational number? It could be. The fact is: we don't know [yet]. It is still an open problem in analysis. As long as someone doesn't prove otherwise, it is very much possible for $\pi+e$ to be a rational number. So, $[1]$ is true.

$[2]$ Notice that, $\pi$ and $e$ are the roots of this equation: $x^2-(\pi+e)x+\pi \times e=0 \cdots (*)$ .

Since, $\pi$ and $e$ are transcendental, they can't be the roots of a polynomial equation with rational coefficients. So, at least one of the coefficients of $(*)$ is irrational. That means at least one of $\pi+e$ and $\pi \times e$ is irrational.

$[3]$ This one's pretty well known. Johann Heinrich Lambert proved that $\pi$ is irrational in the $18$ th century. So, $[3]$ is true as well.

I didn't think $\pi+e$ could be rational. I did not know it was an open problem... Oh well :(

Daniel Liu - 7 years, 2 months ago

Dude, you're not the only one. I feel your pain.

Sharky Kesa - 6 years, 10 months ago

You should take some time to read the Wikipedia page on unsolved problems in mathematics. Some of the problems look so simple that it's hard to believe that they are yet unsolved.

Mursalin Habib - 7 years, 2 months ago

I was careless. i did really know that $\pi + e$ is an open question.

Figel Ilham - 6 years ago

This problem is dangerous! If in the future, someone proved that $\pi+e$ is an irrational number. The [1] will be wrong and [2] should become $\pi\times e$ is irrational. So, this problem can only last for this decade? century? or millennium? just kidding...

Christopher Boo - 7 years, 2 months ago

True, though this problem is equally valid if $\pi$ and $e$ were replaced by $\alpha$ and $\beta$ instead. So not a huge loss.

Calvin Lin Staff - 7 years, 2 months ago

Oh!, then also one must consider the possibility that it is transcendental, we can conclude an answer on the basis of whether its solved or still open!

Krishna Viswanathan - 7 years, 2 months ago

For [3], $\pi$ irrational follows from $\pi$ transcendental.

Patrick Corn - 7 years, 2 months ago

Pie?

1 solution

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