adkikani wrote:
Hi
PKNCan you please explain the highlighted text in
chetan2u 's approach?
Quote:
If a and b are both positive integers such that a < b, which of the following must be true?
Quote:
\(a=\sqrt{aa}\) and b=\(\sqrt{bb}\)since a<b...... \(\sqrt{aa}<\sqrt{ab}<\sqrt{bb}\)......................\(.a<\sqrt{ab}<b\)
I could not proceed from \(\sqrt{a}\) \(\sqrt{b}\) = \(\sqrt{ab}\)
Hi
adkikani ,
First of all, you must be knowing the following property of exponent:-
\((a^m)*(a^n)=a^{m+n}\)
Let \(m=n=\frac{1}{2}\), now \(a^{\frac{1}{2}} * a^{\frac{1}{2}}=a^{\frac{1}{2}+\frac{1}{2}}\)=a----(1)
Also, you know, \(a^{1/2}=\sqrt{a}\)-------(b)
Now moving to the question:-
Given a<b, --------------(c)
I am sure we can write \(a=\sqrt{a}*\sqrt{a}\) & \(b=\sqrt{b}*\sqrt{b}\)
Substituting in (c), we have
\(\sqrt{a}*\sqrt{a} < \sqrt{b}*\sqrt{b}\)-----(d)
Again, a<b (You know we can multiply positive numbers on both sides of the inequality)(Given a and b are positive integers)
Or, \(a*b<b^2\) (multiplying 'b' both sides)
Or, \(\sqrt{ab}<\sqrt{b^2}\) (we can take square root on both sides of inequalities when they are positive)
Or, \(\sqrt{ab}< b\)------------------(e)
Similarly, b>a (You know we can multiply positive numbers on both sides of the inequality)(Given a and b are positive integers)
Or, \(b*a>a^2\) (multiplying 'a' both sides)
Or, \(\sqrt{ab}>\sqrt{a^2}\) (we can take square root on both sides of inequalities when they are positive)
Or, \(\sqrt{ab}> a\)------------------(f)
Combining (d),(e), and (f), we have
\(a<\sqrt{ab}<b\)We have derived it.
_________________
Regards,
PKN
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