Actually, if 0 < a and 0 < b < 1, then ab < a. a does not have to be less than 1.knowLittle said:I would like some feedback on my proof.
Can I just say that :
0< a<1 ... [1]
0<b<1 ... [2]
multiplying [2] by 'a' everywhere, then I get
0<ab<a
And, we prove that ab<a?
To prove this inequality, we can start by noting that since 0 < a < 1, we know that a is a positive number less than 1. Similarly, since 0 < b < 1, we know that b is also a positive number less than 1. Multiplying two numbers less than 1 together will always result in a number smaller than either of the original numbers, hence ab < a.
Considering the range of values for a and b is important in this inequality because it helps us understand the behavior of the variables within a specific range. In this case, when a and b are both between 0 and 1, their product will always be smaller than a. This understanding is crucial in various mathematical and scientific applications.
Sure! Let's take a = 0.5 and b = 0.3 as an example. When we multiply these values together, we get ab = 0.5 * 0.3 = 0.15. Since 0.15 is smaller than 0.5 (a), the inequality ab < a holds true.
The concept of fractions is fundamental in proving this inequality because when a and b are both fractions between 0 and 1, their product will always be a smaller fraction. This is due to the fact that multiplying two fractions results in a smaller fraction, making the inequality ab < a valid.
No, there are no exceptions to this inequality when a and b are both between 0 and 1. The inequality ab < a holds true for all values of a and b within this range, as multiplying two numbers less than 1 will always result in a smaller number than either of the original numbers.