#1**+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'.

GoldenLeaf
Dec 9, 2014

#3**+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'.

GoldenLeaf
Dec 9, 2014

#4**+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

Guest Dec 9, 2014

#6**+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

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

Melody
Dec 9, 2014

#7**+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