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2 Questions, regarding closure

  1. Jan 29, 2008 #1
    1. Show that if [tex]a, b \in \textbf{Q}[/tex], then [tex]ab[/tex] and [tex]a + b[/tex] are elements of [tex]\textbf{Q}[/tex] as well.

    2. Show that if [tex]a \in \textbf{Q}[/tex] and [tex]t \in \textbf{Q}[/tex], then [tex]a + t \in \textbf{I}[/tex] and [tex]at \in \textbg{I}[/tex] as long as [tex]a \neq 0[/tex].

    I'm just a little shady on showing these properties, so if someone could give me a little reminder, that would be swell. It's been a few years since I've taken an Algebra course.
     
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  3. Jan 29, 2008 #2

    HallsofIvy

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    What definitions do you have? With the usual definition of Q: numbers that can be written as a fraction m/n with denominator n not 0, it should be as simple as "if a and b are in Q, then a= m/n, b= p/q for some integers m, n, p, q with m and q not 0. Then a+ b= m/n+ p/q= mq/nq+ np/nq= (mq+ np)/nq- a fraction with numerator and denominator both integers and denominator not 0".

    I assumed that Q mean "the set of rational numbers" because that is the standard notation. I should then assume that I is "the set of all integers" but then question (2) is clearly impossible! The statement you are asked to prove is simply not true. For example, if a= 1/2 and t= 1/3, a+ t= 1/2+ 1/3= 5/6 which is obviously not an integer! Similarly, at= 1/6 which is not an integer.

    Is it possible that you have mis-stated the questions?
     
  4. Jan 29, 2008 #3
    I is the set of irrational numbers. I'm assuming that the second can be proved by contradiction once the first is proven. Essentially, assume that a is rational and t is rational, so when I break it, it's ultimately showing that either a or t is irrational.
     
  5. Jan 29, 2008 #4
    2 is trivially false if both a, and t belong to the rationals. You must mean that exactly one of them belongs to the irrationals in which case it is true.
     
  6. Jan 29, 2008 #5
    It's simple to show rational numbers in terms of ratios, yet there are infinitely many irrational numbers that can't be shown as irrationals. What then would you suggest? I know from the first one that the rationals are closed under addition and multiplication, yet when dealing with irrationals it seems as though it's crossing into unknown territory.
     
  7. Jan 29, 2008 #6
    Did you miss the point of my post entirely? My point is that you say in your first post that both a and t are rational but want to show that a+t and at are irrational. Do you see why this is impossible? So again it must be that exactly one of a and t is irrational in order for 2 to be true.
     
  8. Jan 29, 2008 #7
    the key to these problems is to actually do the algebra.

    if you have a=m/n and b=p/q, actually add and multiply them symbolically to verify for yourself that the results are again rational.

    the second problem is a little different. as d_leet pointed out you meant for t to be irrational, otherwise the statement is false. for this problem a + t = b. if you let b be rational, what absurd conclusion does this imply.
     
  9. Jan 30, 2008 #8
    Well, obviously that t must also be rational, which is again obviously false. Thanks bro!
     
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