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Alternative path to taking roots of both sides of equation

  1. Jul 24, 2015 #1
    The full question is: "How can we take square root of both sides of an inequality or equation just by multiplying each side by numbers with negative rational exponents". I will include several examples to explain how I think about it.
    1)a=b, a^(-0.5)*a=b*a^(-0.5) (but a^(-0.5)=b^(-0.5)) then a^(-0.5)*a=b*b^(-0.5) which is sqrt(a)=sqrt(b)

    2)a>0 b>0
    a>b
    a^(-0.5)*a>b*a^(-0.5) b^(-0.5)*a>b^0.5 ???
    a^(0.5)=b*a^(-0.5)??? (Trying to prove that sqrt(a)>sqrt(b))

    3)x^2=2.5
    x^(-1)*x^2=2.5*2.5^(-0.5) (2.5^(-0.5)=x^(-1))
    x=sqrt(2.5)
    Which is incorrect. Because the true solution is sqrt(x^2)=sqrt(2.5) then |x|=sqrt(2.5) and x1=+sqrt(2.5), x2=-sqrt(2.5)

    (The question has similarities with this one https://www.physicsforums.com/threa...-equation-and-inequality.823960/#post-5174108)
     
  2. jcsd
  3. Jul 24, 2015 #2
    Is what I am doing possible?
     
  4. Jul 24, 2015 #3

    BvU

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    Hello Arman,

    Would it help you to draw some graphs ? The graph for the function ##f(x) = \sqrt x## exists only in the ##f \ge 0 \ {\rm and} \ x \ge 0 ## quadrant. But the graph of ##g(x) = x^2 ## exists in the upper half of the x, g plane. This last one should help you find out why your first step in 3. is a glitch ...
     
  5. Jul 24, 2015 #4
    Hi,
    I think that is because I lost one solution when dividing by x
     
  6. Jul 24, 2015 #5
    I still don't understand 2) it should be true that sqrt(a)>sqrt(b) because it is monotonically increasing
     
  7. Jul 24, 2015 #6

    BvU

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    exactly.

    I don't think you can prove ##a > b \Rightarrow \sqrt a > \sqrt b\ (a>b>0)## this way: it's a bit circular.

    With a bit of liberty, I would do $$ \Bigl ( a> 0 \Rightarrow \sqrt a > 0 \ \& \ b> 0 \Rightarrow \sqrt b > 0 \Bigr ) \Rightarrow \ \sqrt a + \sqrt b > 0 $$ And then use this in
    $$ \Biggl ( a - b > 0 \ \& \ a - b = \left ( \sqrt {\mathstrut a} - \sqrt {\mathstrut b} \right ) \left ( \sqrt {\mathstrut a} + \sqrt {\mathstrut b} \right ) \Biggr )
    \ \ \Rightarrow \ \ \sqrt {\mathstrut a} - \sqrt {\mathstrut b} > 0 $$
     
  8. Jul 24, 2015 #7
    For ##a>b>0,## ##\sqrt{a}>\sqrt{b}## is a trivial implication for ##y=\sqrt{x}## is a strictly increasing function.
     
  9. Jul 25, 2015 #8
    Correct me if I am wrong, but I think your answer is a bit off topic. Anyway thanks for your reply!
     
  10. Jul 25, 2015 #9
    It may be another way to prove ##a>b>0\Rightarrow \sqrt{a}>\sqrt{b}## using the increase of ##y=\sqrt{x}.##
    Let ##f(x)=\sqrt{x},## then ##f(0)=0.## Besides, ##f'(x)=\frac{1}{2\sqrt{x}},## which is always positive when ##x> 0.## As a result, ##f(x)## is a strictly increasing function, so ##a>b>0\Rightarrow \sqrt{a}>\sqrt{b}## is correct.
     
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