Set Problem...........?

Question

S={x|x=3m+2n,m,n∈Z},
T={x|x=5p+4q,p,q∈Z}.
prove that: S=T=R(real number set), and explain WHY?

SORRY SORRY SORRY, !! TYPO,TYPOTYPO.

prove that S=T=Z(integer set)

Answer 1

I believe you want to prove S=T=Z. Clearly, S≠R, since the integers are closed under addition and multiplication, so 3m+2n ∈ Z, but 1/2∈R and 1/2∉Z, so S≠R.

To show that S=Z, we start by showing S⊆Z. This we just did in the above. To show that S⊇Z, we note that ∀k∈Z, ∃m∈Z and n∈Z such that 3m+2n=k (for instance, m=k and n=-k). Therefore, S=Z.

Exactly the same proof works on T (heck, you can even use p=k and q=-k again).

Answer 2

It is not really possible to prove because you have not given us the function Z to which you attribute the domains of S & T. you say nEz and qEz then if S=T, 3m+2n=5p4q. You are also stating that that function S is not continuous,x cannot equal at 3m+2n, and likewise for T you are stating it is not continuous and cannot equal 5p+4q. But then you are stating S=T=R(all reals). The only manner this is possible is if lim(xapproaches 3m+2n)+ of Z(x)=lim(xapproaches 3m+2n)- of Z(x) and likewise lim(xapproaches 5p+4q)+ of Z(x)=lim(xapproaches 5p+4q)- of Z(x). In order to prove S=T=R all you have to do is show that Z is continuous at the points in which both S and T have discontinuities. This will show that Z=S and Z=T but this doesn't necessarily mean S=T unless 3m+2n=5p+4q which is yet another proof you must do for the piecewise function T & S. But I believe this is trully false that Z=S=T because Z is E of all reals, but both S & T have discontinuities and this is not plausible threfore s & T cannot be Elements of reals unless they are directly attributed to the original Z(x) expression.

Answer 3

x E Z --->
x = 3x - 2x --->
x = 3m + 2n where m=x and n=-x --->
x E S.

The converse (x E S ---> x E Z) is obviously true because x=3m+2n, m, n E Z and Z is closed for addition and multiplication.

Ergo S = Z.

You can do the same for T=Z.