Question

A,b) prove that there exists an infinity of rational number

Answer 1

If x≤0 and y≥0, or if x≥0 and y≤0, then z=0 will suffice.

Lemma: ∀x,y∈R with 0<x<y ∃z∈Q such that z∈[x,y].

If x>0 and y>0 and x<y we can apply this lemma directly to show that ∃z∈Q such that z∈[x,y] as required.

If x<0 and y<0 and x<y then −x>0 and −y>0 and −y<−x and so we can apply this lemma to show that ∃z′∈Q such that z′∈[−y,−x]. We can then set z=−z′ and we have z∈Q with z∈[x,y] as required.

If y<x we can exchange x and y and proceed as above.

Thus ∀x,y∈R with x≠y we can find z as required QED

Proof of lemma: ∀x,y∈R with 0<x<y ∃z∈Q such that z∈[x,y].

Since x<y we have y−x>0, so we can pick n∈N sufficiently large that 1n<y−x.

Let k∈Z be the largest integer for which kn≤x. Note that 0n=0<x, so clearly k≥0.

Now let z=k+1n.

z>x, because we chose k to be the largest integer for which kn≤x and k+1>k so we cannot have z=k+1n≤x.

z<y, because k+1n=kn+1n, and we know that kn≤x and 1n<y−x so z=kn+1n<x+(y−x)=y.

z∈Q, since z=k+1n and k+1 and n are both integers.

Thus z∈Q and z∈[x,y] as required QED