How to Prove ( ca geq cb ) Given ( a geq b ) and ( c > 0 )
In this article, we will explore the proof for the inequality ( ca geq cb ) given that ( a geq b ) and ( c > 0 ). This proof will utilize fundamental properties of inequalities and the behavior of positive numbers in the context of real numbers.
Proof Using Inequalities
Let's start with the given inequality:
[ a geq b ]
We are also given that:
[ c > 0 ]
The key property here is that multiplying both sides of an inequality by a positive number preserves the inequality direction. This will be the core of our proof:
Multiply both sides of ( a geq b ) by ( c ).Multiplying both sides of the inequality by a positive number ( c ) results in preserving the inequality direction:
[ ac geq bc ]This completes the proof. The important step is recognizing that the positive number ( c ) does not change the direction of the inequality:
Given: a geq b c > 0Then: ac geq bc
Detailed Considerations
Let's delve deeper into the proof with detailed considerations:
Step 1: We start with the basic inequality ( a geq b ). Step 2: Given that ( c > 0 ), we multiply both sides of the inequality by ( c ): [ c cdot a geq c cdot b ] This step is valid because the multiplication by a positive number ( c ) preserves the inequality.To further solidify this understanding, let's consider the definitions and properties of the sets involved:
Definitions and Properties
Assume ( a, b, c in mathbb{R} ) and ( c in mathbb{P} ), where ( mathbb{P} ) denotes the set of positive real numbers:
By definition, ( a geq b ) implies ( b - a in mathbb{P} ) (set of positive real numbers). Since ( c > 0 ) and the positive real numbers are closed under multiplication, ( c(b - a) in mathbb{P} ). Thus, we have ( c(b - a) geq 0 ), which simplifies to ( cb - ca geq 0 ), or ( cb geq ca ).Alternatively, if we assume that ( ca cb ) and derive a contradiction:
Proof by Contradiction
Assume the statement is false, i.e., ( ca cb ).
Since ( c > 0 ), dividing both sides by ( c ) (which is valid because division by a positive number preserves the equality): [ a b ] But we are given that ( a geq b ) and ( a eq b ), which is a contradiction.Therefore, the original statement must be true:
( ca geq cb )
Thus, we conclude that given ( a geq b ) and ( c > 0 ), the inequality ( ca geq cb ) holds true.