here, there is not much to do after the hint the book gave.
For every positive irrational number b, there exists an irrational number a with 0<a<b
Let b be any positive irrational number. Consider a = b/2. Note that a<b, and since a is obtained by dividing an irrational number by a rational number, a is irrational. And since a is half of a positive number, we have a>0. Thus we have 0<a<b, where a and b are irrational, as desired
Let me show you an example of what I mean. remember this proof?
Let n be an integer. Prove that if n is not congruent to 0 (mod 3), then n^2 is congruent to 1 (mod 3).
We employ a direct proof. We define "n = x (mod 3)" to be n is congruent to x (mod 3). We define "n != x (mod 3)" to be n is not congruent to x (mod 3).
Now assume n != 0 (mod 3). Then n = 1 (mod 3) or n = 2 (mod 3).
Case 1: n = 1 (mod 3)
if n = 1 (mod 3) then by squaring both sides we obtain n^2 = 1^2 (mod 3), that is n^2 = 1 (mod 3)
Case 2: n = 2 (mod 3)
if n = 2 (mod 3) then by squaring both sides we obtain n^2 = 2^2 (mod 3), that is n^2 = 4 (mod 3). Since in a modulus equation, we can add or subtract the modulus as much as we want, this means that n^2 = 4 - 3 (mod 3), that is n^2 = 1 (mod 3).
So we see in both possible cases, n^2 = 1 (mod 3), as desired.
remember, when we say n = 2 (mod 3) for instance we mean, 3|n-2 what that actually means is that if we divide n by 3 the remainder is 2. (do you see why?)
so for mod 3 there are three possibilities:
n = 0 (mod 3)
n = 1 (mod 3)
n = 2 (mod 3)
since when dividing by 3, there can only be three possible remainders, either you have a remainder of 0 (in this case, n is divisible by 3) or you have a remainder of 1 or 2 (in these cases, n is not divisible by 3)
so if i assume n is not congruent to 0 mod 3, that is, it is not divisible by 3, then n has to be one of the other 2
Whey you said, We define "n = x (mod 3)" to be n is congruent to x (mod 3). We define "n != x (mod 3)" to be n is not congruent to x (mod 3). Did you mean n! is not congruent to x? If so that makes sense other wise I don't understand that part. Why did you pick n! = x (mod 3)?
I understand the remainder concept that mod 3 has 3 possibilities.
let's use mod 3 as an example again.
let's say you got n = 4 (mod 3). i claim that is the same as saying n = 4 - 3 (mod 3) which is n = 1 (mod 3)
so now let's think about it. n = 4 (mod 3) means that when we divide n by 3 we get a remainder of 4. but wait, if there is a remeainder of 4, that means we could have gotten another 3 out of it, and be left with a remainder of 1. ah, well if that's true, then n = 1 (mod 3) since we can "take out" a modulus out of 4, it is the same as subtracting the modulus.
so basically, once we go in excess to the point where we can take the modulus out of the remainder, it means we might as well have divided into the remainder more, leaving a smaller remainder. so that's the common sense answer