If a and b are real
If a and b are real numbers. is a<b?
- a^b<b^a
- a/b>1
My Ques: help me solve this by solving equations and not by plug in values please.
Hello Riya,
First things first – the question is on establishing an inequality, so how do you expect to solve it using equations??
Second, one of the most basic methods of solving an inequality is by trying out values and THEN establishing a range that satisfies the given inequality. So, in essence, plugging in values is by no means, an inferior method of solving such problems.
Question data says, a and b are real numbers, which is actually a lot of data. It means, we can take negative integers, negative fractions, negative irrationals, ZERO and the corresponding positive counterparts. Hence, on the number line, we may take FOUR important intervals to solve such questions, which are [0,1], [1,infinity], [0,-1] and [-1,-infinity]. Of course, ZERO would be one of the first values to be checked.
Using statement 1 alone:
If a = 1 and b = 0, a^b < b^a, and also a >b; however, when a = 2 and b = 3, a^b is still lesser than b^a but now, a<b. This is enough to understand that the data given in Statement 1 alone is not sufficient to give a unique answer.
Using statement 2 alone:
a/b>1 is a very generalised statement because this means either a>b (when both positive) or a<b (when both negative). Hence, statement 2 alone is also insufficient.
Now, combining both statements 1 and 2:
As per statement 2, either both a and b have to be positive or both have to be negative.
If both a and b are positive, a^b < b^a and also a>b
If both a and b are negative, for example, if a = -2 and b= -3, a^b < b^a and again a>b.
Hence, we can answer the question on combining data from both statements.
Hope this helps.
Thanks,
Arvind BT