Inequality: If a > b, and if c > d, then....
- Thread starter NSH
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D
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Re: INEQUALITY
That is not a "proper" question - what else is provided here?
NSH said:
That is not a "proper" question - what else is provided here?
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There are any number of "conclusions" one could draw from the above. Without parameters, there is, unfortunately, no possible way of knowing which one particular conclusion is expected for this exercise. Sorry!NSH said:
Please consult with your instructor regarding the missing information. Thank you!
Eliz.
If a>b, and if c>d, then....
If 10>9, and if 6>5, then....you might conclude that 10>5, but...
If 10>9, and if 15>14, then....can you say that 10>14?
If 10>9, and if 6>5, then.... could say that 10+6 > 9+5 or 10-6 > 9-5 or 10*6 > 9*5 or 10/6 > 9/5 ??? One or more of these is not true. One or more are true.
I imagine you are to draw conclusions similar to these, only using a, b, c and d. In other words...
If a>b, and if c>d, then....is a+c always greater than b+d? What if a, b, c, and d are negative numbers?
If 10>9, and if 6>5, then....you might conclude that 10>5, but...
If 10>9, and if 15>14, then....can you say that 10>14?
If 10>9, and if 6>5, then.... could say that 10+6 > 9+5 or 10-6 > 9-5 or 10*6 > 9*5 or 10/6 > 9/5 ??? One or more of these is not true. One or more are true.
I imagine you are to draw conclusions similar to these, only using a, b, c and d. In other words...
If a>b, and if c>d, then....is a+c always greater than b+d? What if a, b, c, and d are negative numbers?
Hello, NSH!
\(\displaystyle \text{If we are concerned with the four arithmetic operations,}\)
. . \(\displaystyle \text{only }addition\text{ is "dependable."}\)
\(\displaystyle \text{We can safely conclude that: }\;\boxed{\:a + c \:> \:b + d\:}\)
\(\displaystyle \text{The others: }\begin{Bmatrix} a-c & >& b-d \\ ac & > &bd \\ \frac{a}{c} & > & \frac{b}{d}\end{Bmatrix}\;\;\text{may or may not be true.}\)
\(\displaystyle \text{If we are concerned with the four arithmetic operations,}\)
. . \(\displaystyle \text{only }addition\text{ is "dependable."}\)
\(\displaystyle \text{We can safely conclude that: }\;\boxed{\:a + c \:> \:b + d\:}\)
\(\displaystyle \text{The others: }\begin{Bmatrix} a-c & >& b-d \\ ac & > &bd \\ \frac{a}{c} & > & \frac{b}{d}\end{Bmatrix}\;\;\text{may or may not be true.}\)
Re:
Thanks.
stapel said:There are any number of "conclusions" one could draw from the above. Without parameters, there is, unfortunately, no possible way of knowing which one particular conclusion is expected for this exercise. Sorry!NSH said:
Please consult with your instructor regarding the missing information. Thank you!
Eliz.
Thanks.
D
Deleted member 4993
Guest
Re: INEQUALITY
This is the only thing that was provided.[/quote]
Then answer is "it cannot be determined".
NSH said:
This is the only thing that was provided.[/quote]
Then answer is "it cannot be determined".