If x and y are real numbers such that 0<x<y^2, then which of the following inequalities must be true?
a. x<y
b. y>0
c. 3x>y^2
d. x<1
e. x<3y^2
Give yourself some values of x and y which make the original statement true.
e.g. x = 2, y = 3 , the 0 < 2 < 9 , <------ true
how about x = .4 , y = .7 , then 0 < .4 < .49 <----- true
how about x = 2, y = -3 , then 0 < 2 < (-3)^2 <----- true
which of the given statements is true for all cases ?
Can you think of any other combinations for x and y which make the original true?
Im confused
To determine which of the given inequalities must be true, let's analyze each option one by one:
a. x < y
Since we know that 0 < x < y^2, we can conclude that x must be less than y because if x were greater than or equal to y, then x would also be greater than or equal to y^2. Therefore, option a is true.
b. y > 0
Since the problem statement only provides the relationship between x and y, it does not give us any information about the sign of y. Therefore, we cannot determine whether y is greater than zero or not based on the given constraints. So, option b cannot be proven true from the given information.
c. 3x > y^2
From the given constraint, 0 < x < y^2, we can multiply each part of the inequality by 3 to get 0 < 3x < 3y^2. However, we cannot conclude that 3x is greater than y^2 because x could be a very small positive number. Therefore, option c is not necessarily true.
d. x < 1
Since we have no information about x other than its relationship with y, we cannot determine whether x is less than 1 or not. Thus, option d cannot be proven true from the given information.
e. x < 3y^2
We know that 0 < x < y^2, so multiplying each part of the inequality by 3, we get 0 < 3x < 3y^2. Therefore, we can conclude that 0 < 3x < 3y^2, which means that x < 3y^2 is true. Hence, option e is true.
In summary, the correct answers are:
a. x < y
e. x < 3y^2