This one seems a little more trivial than the others; I think I see it intuitively but can't really write a more formal proof for it because I feel like I'm skipping a logical step somewhere.
Basically, if you have an irrational number x, you can define a range for x: (a + b/6, a + (b+1)/6), where a is an integer and b is an integer in {0, 1, 2, 3, 4, 5}. It's just a whole number plus a remainder, where the remainder falls into possible pigeonholes between each integer with width 1/6.
Rounding x gets you two values: a and a+1.
Assuming round(x) = a:
For |jx - (a)| < 1/6, you need to be able to show that the remainder term after multiplication by j falls within (0, 1/6). In order for that to happen, you need a value of j where jb ≡ 0 mod 6, which seems trivial because if 6 is a possible value of j, then obviously 6b ≡ 0 mod 6.
Assuming round(x) = a+1:
For |(a+1) - jx| < 1/6, you need to be able to show that the remainder term after multiplication by j falls within (5/6, 6/6), and thus jb ≡ 5 mod 6, which is just as trivial for the possible values of j and b.
Part 2: In the generalized case, if you divide up the number line into segments of width 1/q, there's going to be some integer p such that p/q < e < (p+1)/q. The distance from e to either possible rational bounding it has to be less than 1/q because that's the total length of each interval, and thus |e - p/q| < 1/q.
To make an error less than 1/6, you just define q = 6, which falls within the given bounds (0 < q < 7).
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