1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Very simple logical equation?

  1. Sep 12, 2012 #1
    1. The problem statement, all variables and given/known data
    so this is the question i've trid my best (over 2 hours trynig this and failed) can someone please help me
    show that :
    l a l < 1 and l b l<1 => l a + b l < l 1 + ab l

    2. Relevant equations

    just a little bit on info , you can only stay from the lal < 1 and lbl <1 side to get to the other one

    3. The attempt at a solution
    here's what i tried so fay
    l a l < 1 and l b l<1 => l a b l < 1
    => -1 < ab < 1
    => 0 < 1 + ab <2
    => l 1 + ab l <2
    l a l < 1 and l b l<1 => -2 < a + b < 2
    => l a + b l < 2
    yup im pretty much stuck here..pleasee help(i can't say l a + b l < l 1 + ab l because they are both<2..
     
  2. jcsd
  3. Sep 12, 2012 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Don't worry too much about staying on one side while you figure out what's going on. You can always modify your argument to fix that later. You can even work backwards while analyzing it. So far you have 0 < 1+ab, so your problem is equivalent to |a+b|<1+ab. Write that as$$
    \sqrt{(a+b)^2} < 1+ab$$Square both sides and work on that. Once you see it, modify it by working backwards through the reversible steps.
     
  4. Sep 12, 2012 #3
    if |a|=0.8 which is less than 1 and |b|=0.2 which is also less than 1 but |ab|= 1.6 which is greater than 1. so your very first line of the solution is wrong
     
  5. Sep 12, 2012 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    |a|+|b| ? |a + b|
    |a||b| ? |ab|
    As LCKurtz says - dont worry about the rules while you are trying to understand the relations.

    @KUMAR |0.8x0.2| = 0.16 < 1 isn't it?
     
  6. Sep 12, 2012 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    ##(.2)( .8) = .16##
     
  7. Sep 12, 2012 #6
    thank you
     
  8. Sep 12, 2012 #7
    tried this solution?

    let us assume that
    |a+b| < | 1+ab|
    squaring both sides
    a*a + 2ab + b*b < 1 + 2ab + a*a*b*b ----- 1

    we have
    |a |< |1|
    square both side
    a*a < 1
    multiply with b*b it is positive and no change in inequality
    a*a*b*b < b*b
    adding 1 on both sides and 2ab
    1 + 2ab +a*a*b*b < 1 + 2ab + b*b ---- 2
    using 1 and 2
    we have
    a*a + 2ab + b*b < 1+ 2ab + b*b
    which boil down to
    a*a < 1 which is true
    I do not know whether this approach is right or wrong
     
  9. Sep 12, 2012 #8

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I would call that an exploratory argument. It shows you why the inequality works so you understand it. But to make a correct argument, you must start with something that is true and end up with your inequality. So your argument would start like this:

    |a| < 1 is given
    |a|2 = a2 < 1 ...

    Continue by reversing your steps and ending up with what you were to prove.
     
  10. Sep 12, 2012 #9
    thanks again i think i got it :)
     
  11. Sep 12, 2012 #10
    Hint:
    If [itex]a = 0[/itex] or [itex]b = 0[/itex], then the identity is surely satisfied (why?).

    If [itex]a \neq 0[/itex] and [itex]b \neq 0[/itex], then [itex]\vert a \, b \vert > 0[/itex]:
    Apply the triangle identity for the absolute value to:
    [tex]
    \left\vert \frac{1}{a} + \frac{1}{b} \right\vert
    [/tex]
    and multiply both sides of the inequality by [itex]\vert a \, b \vert[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Very simple logical equation?
  1. Very simple lim (Replies: 4)

  2. Very simple ODE (Replies: 6)

Loading...