[1] You wouldn't. It's not true: the square root of 4 is rational.
[2] I think this proof follows pretty much the same lines as for the square root of 2.
I know to prove the irrationality of we simply minipulate the following:
until we hit the contradiction of not being in reduced terms. I also know that we must take into account that all number can be written in the form or for ; and the fact that if is an even number then is an even number. Heres my question, how do you prove the irrationality of a certain opertation on numbers in general? And for odd square roots?
Two examples of what I mean:
[1] How would you prove that if that is always irrational?
[2] How would you prove is irrational? Would you find a way to define numbers in terms of 3's ? Just like with the ? So all numbers would be of the form , , and ; and would you some how use that fact to prove the irrationality of ?
Thanks in advance
[1] You wouldn't. It's not true: the square root of 4 is rational.
[2] I think this proof follows pretty much the same lines as for the square root of 2.
For (2), it's the same way how you would prove the irrationality of root(2).
Assuming that is rational, then it takes the form , where and m and n do not have common factors.
Since is an integer, it follows that m/3 is an integer too. Thus, 3 is a factor of m.
Let where ,
We see that both m and n have a common factor of 3 which contradicts our assumption.
Okay, your right, I think I've figured out the proof for all even numbers, but I'm alittle lost for the odd numbers. Here is what I have, please point out any mistakes I made.
-> We want to prove that is irrational, as long as no numbers satisfy .
[i] If is rational, then it can be written in the following form:
thus
[ii] This leads to the following equation:
is either even or odd. Lets start with even.
[Case 1]
Say is even, then for some we have
Thus, if is even, [of the form ]
then is even, and of the form
for some natural number .
Thus, we have and :
This proves to be even.
If is even, then we'd arrive at the form (for some )
thus,
for some
Thus and have a common divisor, and this is a contradiction.
Is this correct? For this one case where for some and for some ? I don't know exactly how to go about proving it with the odd numbers, I tried on pen and paper using the form for but the "common factor" never seemed to show up. Can somebody give me a hint or show me the right direction?
When you use the Unique Factorization Theorem, you get a simple proof.[1] How would you prove that if that is always irrational?
Let be a positive integer. We can show that if is rational, then must be a perfect square. From here it follows that if is not a perfect square, then is irrational.
Suppose that for positive. Then . For any prime , let and be the highest powers of that divide and , respectively.
Now, because and using the Unique Factorication Theorem, we must have and then . This means that is a perfect square.
This way you can even prove a more general result: For any positive integers ; if is rational, then must be a perfect th power.
There is a nice proof I saw once. It can be generalized to any positive integer that is not a perfect sqaure.
Soppose that , where are postive integers that are relatively prime. Then for some integers . (Bézout's Identity)
Because , we have and . Then , and is clearly an integer.
This means that is an integer, but ; contradiction.