Math Is Fun Forum

dchilow

Member

Registered: 2007-03-05

Posts: 27

Can anyone help with Real Analysis?

Let  a  be a nonzero real number and let S be the set:  S = {x∈R: |x-a| < |a|/2}

(a)  Show that |x|>|a|/2 for all x∈S.

(b)  Write S as either an interval or a union of intervals.

Please help me on this, thanks!

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Can anyone help with Real Analysis?

Any time you see |a - b|, you should think of this as the distance between a and b, as that is what it really is.

|x-a| < |a|/2

We have that this distance between x and a is less than |a|/2.  If |x| <= |a|/2, then what does this mean about the distance between x and a?  If you need to, separate it into four cases with x and a being either positive of negative.  But you can get it all in one shot if you approach it right.

Viewing |x-a| as a distance, it should be rather immediate what the interval will be.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Can anyone help with Real Analysis?

if a is negative, then S is between 3a/2 and a/2.
if a is positive, then S is also between the other 3a/2 and a/2.

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igloo myrtilles fourmis

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Can anyone help with Real Analysis?

As I was trying to explain, there is no need to do it case wise.

The interval is (a - |a|/2, a + |a|/2)

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Can anyone help with Real Analysis?

Sorry that was just for clarity, nothing more.

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igloo myrtilles fourmis