Question

Prove\[\dfrac{{a + x}}{{b + x}} < \dfrac{a}{b}\] , if $a > b$ and x is $ + ve$ .

Answer 1

Hint: In this problem, we have given that a is greater than b $(a > b)$, means the value of a is greater than b and here is also given that x is positive $( + ve)$ and with the help of this, we have to prove that the given condition \[\dfrac{{a + x}}{{b + x}} < \dfrac{a}{b}\] is also true. This can also be proved by assuming a, b and x.

Complete step by step answer:
To prove the above condition, let us assume that the given condition \[\dfrac{{a + x}}{{b + x}} < \dfrac{a}{b}\] is true.
Now, we will multiply the whole condition with $b + x$,
$ \Rightarrow \dfrac{{a + x}}{{b + x}} \times (b + x) < \dfrac{a}{b} \times (b + x)$
On further solving we get,
$ \Rightarrow (a + x) < \dfrac{{a(b + x)}}{b}$
Now, we will multiply the whole condition with b,
$ \Rightarrow (a + x) \times b < \dfrac{{a(b + x)}}{b} \times b$
On further solving we get,
$ \Rightarrow b(a + x) < a(b + x)$
(Where, a, b are positive.)
On simplifying,
$ \Rightarrow ab + bx < ab + ax$
Now, subtract $ab$ from both sides,
$
   \Rightarrow ab + bx - ab < ab + ax - ab \\
   \Rightarrow bx < ax \\
 $
Divide by x on both sides,
$ \Rightarrow \dfrac{{bx}}{x} < \dfrac{{ax}}{x}$
(x is positive)
On further solving, we get,
$ \Rightarrow b < a$
Which is also given in our question, hence the condition \[\dfrac{{a + x}}{{b + x}} < \dfrac{a}{b}\] is proved by the help of the given conditions i.e, $a > b$ and x is positive.

Note: We have proved the given condition \[\dfrac{{a + x}}{{b + x}} < \dfrac{a}{b}\] is true by firstly assuming it true, and after solving it we reached to the condition that is given in the question i.e.$a > b$, which proves that the given condition is true. We can also prove this condition by assuming$a = 2$, then b should be less than a$(a > b)$, then we assume $b = 1$, and x is positive, so we assume$x = 3$. Now, let us substitute the values of a, b and x into the condition,
$ \Rightarrow \dfrac{{2 + 3}}{{1 + 3}} < \dfrac{2}{1}$
On further solving, we get,
$
   \Rightarrow \dfrac{5}{4} < 2 \\
   \Rightarrow 1.25 < 2 \\
 $
Hence, proved that the given condition is true.