2) Prove these by induction.
a) Inductive step: 2n+3 = (2n+1) + 2 < 2^n + 2 ≤ 2^n + 2^n = 2^(n+1).
b) Inductive Step: n+1 < 2^n + 1 ≤ 2^n + 2^n = 2^(n+1).
3) Write n = 2k+1 for some integer k.
2^(2k+1) + 3^(2k+1)
= 2 * 4^k + 3 * 9^k
= 2 * 4^k + 3 * (4 + 5)^k
= 2 * 4^k + 3 * [4^k + (terms divisible by 5)], via binomial theorem
= 5 * 4^k + 3 * (terms divisible by 5).
4) If f(x) = f(x'), then 3x - 7 = 3x' - 7 ==> x = x'; hence f is 1-1.
f(R) = R, since for any y in R, f((y+7)/3) = y.
5) Note that Q is countable, there exists a bijection f: Q --> N.
So, we have a bijection between Q^3 and N^3 via g: Q^3 --> N^3 via g(a, b, c) = (f(a), f(b), f(c)).
To show that N^3 is countable, note that we have an injection
h: N^3 --> N defined by h(a,b,c) = 2^a * 3^b * 5^c.
(h is a bijection, due to the Fundamental Theorem of Arithmetic).
Therefore, Q is countable.
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I hope this helps!