Why does sqrt(24) = 2*sqrt(6)?

[nickadams]

I realize the process for simplifying square roots is to break it down into lowest factors and then bring the pairs out of the radical, but why does that work? Can anybody offer an explanation?

I know it's a really elementary question but i never understood why it works!

 

[iRaid]

\sqrt{24}=\sqrt{(4)(6)}
Square root of 4? = 2...
2\sqrt{6}=\sqrt{24}

 

[nickadams]

\sqrt{24}=\sqrt{(4)(6)}
Square root of 4? = 2...
2\sqrt{6}=\sqrt{24}

ya i get that, but i don't understand why it works. To me it's just a little trick that works but i don't know why

 

[Deveno]

I realize the process for simplifying square roots is to break it down into lowest factors and then bring the pairs out of the radical, but why does that work? Can anybody offer an explanation?

I know it's a really elementary question but i never understood why it works!

well, it all hinges on this fact:

for a,b ≥ 0, √(ab) = (√a)(√b).

how can we prove this? well, we can just square (√a)(√b) and see if we wind up with ab.

to do THAT, we need another fact (which is probably easier for you to swallow):

(xy)² = x²y²

because: (xy)² = (xy)(xy) = x(yx)y = x(xy)y = (xx)(yy) = x²y².

so...[(√a)(√b)]² = (√a)²(√b)² = ab

(by the definition of square root), and since a,b ≥ 0, √a, √b ≥ 0 (we are only interested in the POSITIVE square root), and positive (to be truly accurate, non-negative) times positive is positive (if one of a,b (or both) is 0, then ab = 0, and the square root of 0 is 0).

therefore, (√a)(√b) is indeed the positive square root of ab.

 

[nickadams]

well, it all hinges on this fact:

for a,b ≥ 0, √(ab) = (√a)(√b).

how can we prove this? well, we can just square (√a)(√b) and see if we wind up with ab.

to do THAT, we need another fact (which is probably easier for you to swallow):

(xy)² = x²y²

because: (xy)² = (xy)(xy) = x(yx)y = x(xy)y = (xx)(yy) = x²y².

so...[(√a)(√b)]² = (√a)²(√b)² = ab

(by the definition of square root), and since a,b ≥ 0, √a, √b ≥ 0 (we are only interested in the POSITIVE square root), and positive (to be truly accurate, non-negative) times positive is positive (if one of a,b (or both) is 0, then ab = 0, and the square root of 0 is 0).

therefore, (√a)(√b) is indeed the positive square root of ab.

thanks that was great! But can you explain how you did...

(xy)(xy) = x(yx)y

?

what is that step? I knew (xy)(xy) = x^2y^2 but i don't get your intermediate steps... The rest was crystal clear thank you

 

[iRaid]

x(yx)y is just multiplying, you can multiply in any order and you will always get the same number:

Example: 2*3*4*5 = 5*3*4*2 = 4*5*3*2 = etc..

 

[mathwonk]

square both sides. you get the same thing. then what?

 

[nickadams]

x(yx)y is just multiplying, you can multiply in any order and you will always get the same number:

Example: 2*3*4*5 = 5*3*4*2 = 4*5*3*2 = etc..

oh that's true. Thanks for the help iRaid

square both sides. you get the same thing. then what?

24=4*6... so they're equal lol

thanks everyone

 

[Deveno]

thanks that was great! But can you explain how you did...

(xy)(xy) = x(yx)y

?

what is that step? I knew (xy)(xy) = x^2y^2 but i don't get your intermediate steps... The rest was crystal clear thank you

in more complete detail:

(xy)(xy) = ((xy)x)y = (x(yx))y = x(yx)y

the re-positioning of the parentheses is justified by the associative law for multiplication:

for any 3 (real, rational, integer) numbers a,b,c:

(ab)c = a(bc).

associativity is often over-looked as a property of multiplication because we just take it for granted and often write:

abc, for any triple (or longer) product.

so i could have just written:

(xy)²= x*y*x*y = x*x*y*y = x²y²

(where all i did is "switch" the middle two factors, since multiplication for (real, rational, integer) numbers is also commutatitive).