Conditions for which inequalitys is true

  • Thread starter Thread starter transgalactic
  • Start date Start date
  • Tags Tags
    Conditions
transgalactic
Messages
1,386
Reaction score
0
A:
1-mt<w^2(m+t)

B:
-(1+mt)<(w^2 )t(m-t)


i have 4 cases
for each case they said that needs to be a condition for which this case would be true
1: m>t and mt<1
so they say that i have to demand
w^2>(1-mt)/(m+t) which is inequality A

2: m<t and mt<1
they say that i have to demand
inequalitys A and B

why ??
 
Physics news on Phys.org
transgalactic said:
A:
1-mt<w^2(m+t)

B:
-(1+mt)<(w^2 )t(m-t)


i have 4 cases
for each case they said that needs to be a condition for which this case would be true
1: m>t and mt<1
so they say that i have to demand
w^2>(1-mt)/(m+t) which is inequality A

2: m<t and mt<1
they say that i have to demand
inequalitys A and B

why ??
If m< t, m- t is negative, if m> t, m- t is positive, if mt< 1, 1-mt is positive and w^2 is always positive.
 
so why in case two they demand both ??
it seems like the left side is true always
it cannot be negative?
 
why in the 1st case we only pick only one innequality
??
 
even if i substitute what you say regarding the positive and negative
i get
positive>positive

so??
 
wwoowww thanks i understood that stuff
:)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top