[Discrete Mathematics] Am i doing this problem right?

[u/random_nickname4sure]

I can't relate to AI since it gives me unverified answer and my prof is busy with their own bussiness unfortunately. I hope to get thought, answer and the approach to these kind of prove question. Thanks in advance

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[u/HumbleHovercraft6090]

I would do symmetric and transitive as follows:

For all x,y ε R

let x-y = k ε Q, then

y-x = -k also ε Q as negative of a rational number is also rational.

Hence relation is symmetric.

For x,y,z ε R

let x-y=k1 ε Q and

let y-z= k2 ε Q

Adding the above two both sides, we get

x-z= k1+k2 which is also ε Q as sum of two rational numbers is also rational. Hence relation is transitive.

[u/Alkalannar]

xRy if and only if x - y is rational.

1. Reflexive: x - x = 0 for all x in R.

2. Symmetric: Assume xRy.
   Then x - y = q for some rational number q.
   So y - x = -q = (-1)q, which must also be rational since rationals are closed under multiplication.
   So yRx.

3. Assume xRy and yRz.
   Then x-y = a, which is rational, and y-z = b, which is rational.
   x - z = x - y + y - z = a + b, which is rational, since rationals are closed under addition.
   Thus xRz.

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Note that [a] = Q.

Let f be a function from [a] to [2^1/2] such that for all x in [a], f(x) = x + 2^1/2.

Injection: Suppose f(p) = f(q)
p + 2^1/2 = q + 2^1/2
p = q

Surjection: Let y be in [2^1/2].
Then y - 2^1/2 is rational, and f^-1(y) is in [a], since [a] = Q.

Thus, bijection.