Thread: Question about a question relating to irrationality.

So there is a problem here: mathschallenge.net

That asks a person to show that e is irrational. Well, I just thought that since (for example) ...8281.. was in the string of decimals for e it had to be irrational because I was taught in elementary school that irrational numbers are those that don't repeat and go on forever and you can find one by seeing if a number, such as 8, is repeated but, followed by a different number (1 the second time instead of 2). Was I taught wrong as in, did they simplify stuff for me in elementary school? Because the solution they give is a lot more complicated than that.

Originally Posted by thyrgle

So there is a problem here: mathschallenge.net

That asks a person to show that e is irrational. Well, I just thought that since (for example) ...8281.. was in the string of decimals for e it had to be irrational because I was taught in elementary school that irrational numbers are those that don't repeat and go on forever and you can find one by seeing if a number, such as 8, is repeated but, followed by a different number (1 the second time instead of 2). Was I taught wrong as in, did they simplify stuff for me in elementary school? Because the solution they give is a lot more complicated than that.

It is true that that irrational numbers do not have repeating blocks of digit
for example
the group of digits repeat that mean the number is rational

Note that the repeating does not have to start at the beginning.

Basically an irrational number will never stay repeating any block of digits.

@TheEmptySet: Thanks for the reply, but the website also has a proof too, but why is my way wrong?

Originally Posted by thyrgle

@TheEmptySet: Thanks for the reply, but the website also has a proof too, but why is my way wrong?

Sorry I misread your original question. Using your defintiion you would have to show that no matter how many decimal places to look out a million, billion, trillion ect that a pattern didn't emerge. That creates a huge problem. How many digits dig you look at 10,50,100 it could happen much much after that. That is why that method wont work. The proof above bypasses that problem by using a proof by contradiction.