Recent content by ArmanZ

Correct me if I am wrong, but I think your answer is a bit off topic. Anyway thanks for your reply!

I still don't understand 2) it should be true that sqrt(a)>sqrt(b) because it is monotonically increasing

Hi, I think that is because I lost one solution when dividing by x

Is what I am doing possible?

The full question is: "How can we take square root of both sides of an inequality or equation just by multiplying each side by numbers with negative rational exponents". I will include several examples to explain how I think about it. 1)a=b, a^(-0.5)*a=b*a^(-0.5) (but a^(-0.5)=b^(-0.5)) then...

For everyone who found this discussion, in this comment I was implying 2<4, 4<16⇔2*2<4*4, 2*2<4*4. But anyway, I hope the general idea of this discussion is understood correctly by everyone.

Sorry, I made a careless mistake once again

I think my question is not clear. For example 2>4(both are positive).If we square both sides we get 4>16, which would be correct if we did did not consider the following fact: 4>16⇔2*2>4*4 but how could we get 2*2>4*4 from 2>4 if we have to multiply both sides by different numbers( 2 and 4, 2≠4)...

What is a square of a number? A^2=A*A. If A=B squaring both sides will give A^2=B^2. How I think about squaring is we multiply both sides of A=B by A(we could also do this for B) we get A*A=B*A but A=B so this will result in A*A=B*B. But if we do this for an inequality, A>B, multiplying both...