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?
Yes, your proof is correct. Since both a and b are positive numbers less than 1, their product ab will also be less than a. This is because multiplying by a fraction (a number between 0 and 1) reduces the value of a.
The inequality ab < a holds true because multiplying a number a (where 0 < a < 1) by another number b (where 0 < b < 1) results in a product that is smaller than a. This is due to the properties of fractions: multiplying by a fraction reduces the value.
Sure, consider a = 0.5 and b = 0.6. Both numbers are between 0 and 1. The product ab = 0.5 * 0.6 = 0.3, which is indeed less than 0.5 (the value of a). This demonstrates that ab < a.
Yes, the inequality ab < a is always true for any values of a and b between 0 and 1. This is because both a and b are positive fractions, and their product will always be a smaller fraction, thus less than a.
To formally prove that ab < a given 0 < a < 1 and 0 < b < 1, consider that since b is between 0 and 1, we can write b as b = 1 - c where 0 < c < 1. Then ab = a(1 - c) = a - ac. Since ac is positive (because both a and c are positive and less than 1), it follows that a - ac < a, hence ab < a.