+0  
 

Best Answer 

 #7
avatar+33603 
+13

If a and b are both negative then 

$$\sqrt a\times \sqrt b=\sqrt{-|a|}\times \sqrt{-|b|}=i^2\sqrt{|a|\times |b|}=-\sqrt{|a| \times |b|}$$

But

 

 $$\sqrt{a\times b}=\sqrt{-|a|\times -|b|}=\sqrt{|a|\times|b|}$$

i.e. the two are different, so not true when both are negative.

.

 Dec 9, 2014
 #1
avatar+1006 
+5

Been a while since I've toyed around with imaginary numbers, but I'm pretty sure that, staying on the real side of numbers and away from imaginary numbers, the domain of sqrt(a) * sqrt(b) equals [a > 0; b > 0], or all positive numbers for both 'a' and 'b'.

 Dec 9, 2014
 #2
avatar
+5

a and b can be 0 tho

 Dec 9, 2014
 #3
avatar+1006 
+5

Yes Anon, a and b can equal 0. I blame my computer not having the 'greater than or equal to' keys >.<

 

But no numbers below 0, basically no negative numbers can equal 'a' OR 'b'.

 Dec 9, 2014
 #4
avatar
+5

a and b can be negative number 

x^2+1=0

      x^2=-1

          x = positive or negative square root of -1

In the 1500s .,an italian mathematician named rafael bombelli invented the imaginary number , which is now called i.  Square root = i implie that i^2=-1 after his invention , it became possible to find soloution for x^2+1=0: they are i and -i. The value of square root of -16 =square root of 16 *(-1)=square root of 16*i^2=4i  

bUt the domain of a and b in that formula  is a>or=0 and b>or =0

 Dec 9, 2014
 #5
avatar
0

no one anwser my original anwser yet! I need more help!

 Dec 9, 2014
 #6
avatar+118587 
+8

$$\sqrt{a}+\sqrt{b}=\sqrt{ab}$$

There is no domain here - the word does not make sense here - not to me anyway.

 

If you are only considering real numbers then a and be will be restricted as follows

$$a\ge 0 \qquad and \qquad b \ge 0$$

.
 Dec 9, 2014
 #7
avatar+33603 
+13
Best Answer

If a and b are both negative then 

$$\sqrt a\times \sqrt b=\sqrt{-|a|}\times \sqrt{-|b|}=i^2\sqrt{|a|\times |b|}=-\sqrt{|a| \times |b|}$$

But

 

 $$\sqrt{a\times b}=\sqrt{-|a|\times -|b|}=\sqrt{|a|\times|b|}$$

i.e. the two are different, so not true when both are negative.

.

Alan Dec 9, 2014
 #8
avatar
+5

Thank you guys!

 Dec 9, 2014

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