Alternative path to taking roots of both sides of equation

[ArmanZ]

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)

 

[ArmanZ]

Is what I am doing possible?

 

[BvU]

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 ...

 

[ArmanZ]

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 ...

Hi,
I think that is because I lost one solution when dividing by x

 

[ArmanZ]

I still don't understand 2) it should be true that sqrt(a)>sqrt(b) because it is monotonically increasing

 

[BvU]

I think that is because I lost one solution when dividing by x

exactly.

it should be true that sqrt(a)>sqrt(b) because it is monotonically increasing

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
$$ \Biggl ( a - b > 0 \ \& \ a - b = \left ( \sqrt {\mathstrut a} - \sqrt {\mathstrut b} \right ) \left ( \sqrt {\mathstrut a} + \sqrt {\mathstrut b} \right ) \Biggr )
\ \ \Rightarrow \ \ \sqrt {\mathstrut a} - \sqrt {\mathstrut b} > 0 $$

 

[tommyxu3]

For $a>b>0,$ $\sqrt{a}>\sqrt{b}$ is a trivial implication for $y=\sqrt{x}$ is a strictly increasing function.

 

[ArmanZ]

For $a>b>0,$ $\sqrt{a}>\sqrt{b}$ is a trivial implication for $y=\sqrt{x}$ is a strictly increasing function.

Correct me if I am wrong, but I think your answer is a bit off topic. Anyway thanks for your reply!

 

[tommyxu3]

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.