Terry W. answered • 02/20/15

Experienced Tutor Specializing in STEM Subjects

So basically you have 2 sets of numbers that satisfies the equations 4s-1 and 2t-1. Since you are given the condition that s and t must be integers you know that the 2 sets must also only contain integers. So what the question is asking is that is every member of set A (4s-1) also a member of set B (2t-1)?

So here's a simple take on the problem.

So here's a simple take on the problem.

Let's assume that Set A and Set B have some common members (a less stringent condition than what you need to prove). In that case, for the intersection of the 2 sets you can set the 2 equations equal to each other and solved for t:

4s-1=2t-1

4s=2t

t=2s

4s-1=2t-1

4s=2t

t=2s

So what does this tell you? We know that 2*integer will always be an integer so if s is always an integer then t must also always be an integer. Notice however that the reverse is not true. Half of an integer is not always an integer (ie 3/2=1.5) so we know that t being an integer does not necessarily constrain s to also be one.

Based on the above, the intersection of set A and set B can now be rewritten in set A terms using s=t/2:

set A {x|x=4(t/2)-1=2t-1 where t=2s and s is an integer}

In other words:

set A {set B|t=2s for s is an integer}

Since this condition includes all of set A, this shows that set A is a subset of set B.

Sorry if the logic is a bit convoluted here.

Edit:

I'll note that doing it the other way is much simpler:

You can rewrite the intersection of Set A and Set B in terms of Set B using the relation t=2s:

A int B = set B {x|x=2(2s)-1 for s is an integer} = set B {x|x=4s-1 for s is an integer} = set B {set A}

Thus you've proven that set A is entirely contained within the intersection of set A and set B (ie set A is a subset of set B)