Is My Proof That ab<a Correct Given 0<a<1 and 0<b<1?

[knowLittle]

I would like some feedback on my proof.

Can I just say that :
0< a<1 ... [1]
0<b<1 ... [2]

multiplying [2] by 'a' everywhere, then I get
0<ab<a

And, we prove that ab<a?

 

[.Scott]

Yes. But you need to note that the multiply does not reverse the inequality signs because "a" is always positive.
(and don't you want a final QED?)

 

[topsquark]

I would like some feedback on my proof.

Can I just say that :
0< a<1 ... [1]
0<b<1 ... [2]

multiplying [2] by 'a' everywhere, then I get
0<ab<a

And, we prove that ab<a?

Actually, if 0 < a and 0 < b < 1, then ab < a. a does not have to be less than 1.

-Dan

 

[WWGD]

I don't want to be overkill on this, but the proof will depend on the axioms and rules you're going by.
You may, though, move the a to the other side and factor.

 

[knowLittle]

Thank you all!