Proof of \sqrt{ab} = \sqrt{a}\sqrt{b} for all ab>0

[annoymage]

Homework Statement

$$\sqrt{ab}$$ = $$\sqrt{a}$$$$\sqrt{b}$$ for all ab>0

Homework Equations

The Attempt at a Solution

let a=0, $$\sqrt{0}$$ = $$\sqrt{0}$$

assume that it's true for some a, consider a+1

$$\sqrt{(a+1)b}$$ = and I'm lost

 

[Mark44]

Homework Statement

$$\sqrt{ab}$$ = $$\sqrt{a}$$$$\sqrt{b}$$ for all ab>0

Homework Equations

The Attempt at a Solution

let a=0, $$\sqrt{0}$$ = $$\sqrt{0}$$

assume that it's true for some a, consider a+1

$$\sqrt{(a+1)b}$$ = and I'm lost

Why do you think that this should be done by induction?

 

[annoymage]

ooops, yea, on second thought, induction is for integer, not real numbers

is there any ways to proof this?

 

[annoymage]

but wait,

$$\sqrt{ab}$$ = $$\sqrt{a}$$$$\sqrt{b}$$ for all integer a and b such that ab>0

how about that? can i solve this using induction?

 

[Mark44]

How is the problem stated? Does it explicitly specify that a and b are integers? Also, does it say ab > 0 or does it say a > 0 and b > 0? There's a difference.

 

[annoymage]

How is the problem stated? Does it explicitly specify that a and b are integers? Also, does it say ab > 0 or does it say a > 0 and b > 0? There's a difference.

no, i just made one up.

but what about

$$\sqrt{ab}$$ = $$\sqrt{a}$$$$\sqrt{b}$$ for all integer a and b such that ab>0

can't i prove it using induction?

 

[Mark44]

Induction really isn't a good fit here. With an induction proof you have a sequence of statements P(1), P(2), P(3), ..., P(n), ... With the problem you made up, you have two variables a and b, both of which should be real and nonnegative.