cubicmonkey said:I'm completely lost on this one. I need this to be able to solve that the square root of three is irrational. So it's a proof within a proof, but I like this way best. Please help me out.
This is what I need to know how to prove.
3|n^2 implies 3|n where n is some integer.
"3|n^2" means that 3 is a divisor of n^2, or in other words, n^2 is a multiple of 3.
If 3 is a divisor of n^2, then n^2 can be written as 3k for some integer k. This means that n^2 is a multiple of 3, and therefore n must also be a multiple of 3 in order for n^2 to be a multiple of 3. Hence, "3|n^2" implies "3|n".
Yes, for example, n = 6. 3 is a divisor of 6^2 (36), but 3 is not a divisor of 6.
This statement is essentially a divisibility rule for 3. It states that if a number is divisible by 3, then its square is also divisible by 3.
No, "3|n^2" is not a sufficient condition for "3|n". It only tells us that if "3|n^2" is true, then "3|n" must also be true. However, there may be other factors that contribute to "3|n" being true, such as n itself being a multiple of 3.