Show $\frac{3997}{4001}>\frac{4996}{5001}$

Question

I wish to show that $$\frac{3997}{4001}>\frac{4996}{5001}.$$

Of course, with a calculator, this is incredibly simple. But is there anyway of showing this through pure analysis? So far, I just rewrote the fractions:

$$\frac{4000-3}{4000+1}>\frac{5000-4}{5000+1}.$$

Answer 1

Continuing with your approach, write

$\frac{4000(1-3/4000)}{4000(1+1/4000)}>\frac{5000(1-4/5000)}{5000(1+1/5000)}$

Then cancel on both sides and cross multiply:

$(1-\frac{3}{4000})(1+\frac{1}{5000})>(1+\frac{1}{4000})(1-\frac{4}{5000})$

Now, expand both sides:

$1+\frac{1}{5000}-\frac{3}{4000}-\frac{3}{20000000}>1+\frac{1}{4000}-\frac{4}{5000}-\frac{4}{20000000}$

$\frac{5}{5000}-\frac{4}{4000}+\frac{1}{20000000}>0$

$\frac{1}{20000000}>0$.

Since the above is clearly true, then the original statement is true also.

Answer 2

Subtract $1$ from both sides:

$$-\frac{4}{4001} > -\frac{5}{5001}$$

Remove the negatives and flip the inequality:

$$\frac{4}{4001} < \frac{5}{5001}$$

This is equivalent to

$$\frac{1}{1000.25} < \frac{1}{1000.2}$$

Multiply both sides by $1000.25 \cdot 1000.2$ to get

$$1000.2 < 1000.25$$

Since this is true, the original inequality must be true.

Answer 3

For positive real $a,b,c,d$, $\frac ab > \frac cd$ if and only if $ad - bc > 0$.

So cross multiplication is the key.

What you should compute is $(4000-3)(5000+1) - (5000-4)(4000+1)$.

Expand to get:

$(4000)(5000) - 3(5000) + 1(4000) - 3 - (5000)(4000) + 4(4000) - 1(5000) + 4$

and note that $(4000)(5000) - (5000)(4000) = 0$ (cancellation)

leaving you with:

$ - 3(5000) + 1(4000) + 4(4000) - 1(5000) + 1$

after a little simplification, and which becomes:

$5(4000) - 4(5000) + 1$

after some grouping of terms and it's trivial to see that that becomes $1$.

Answer 4

This is similar to some other posts, but I find it a little simpler and easier, and does not have fractions.

$$\frac {4000-3} {4000+1} \gt \frac {5000-4} {5000+1}$$

Multiply by $({4000+1})*({5000+1)}$

$$({4000-3})*({5000+1}) \gt ({5000-4})*({4000+1})$$

Subtract 20,000,000 ($5000*4000$)
Subtract 5000
Subtract 4000

$$-3-(4*5000) \gt -4-(5*4000)$$

Add 20000 $$-3 \gt -4$$

Answer 5

First a minor but cute detour.

$a_n=\frac {n}{n*1000 +1} = \frac 1{1000 + 1/n} $.

As $n $ gets larger $1/n $ gets smaller, so $a_n $ get larger. So $a_1 < a_2 < a_3 <.... $.

So $\frac {3997}{4001} = 1 - \frac 4 {4001} = 1 - a_4 > 1 - a_5 = 1 - \frac 5 {5001}=\frac {4996}{5001} $.

Okay, this wasn't as straightforward or as easy as the others but in my opinion it was i) informative as why it ought to be true and ii) cute.

Answer 6

Using long multiplication we get $$3997\times5001=19988997>19988996=4996\times4001$$ which implies the desired result (because positive multiplication preserves order).

Answer 7

For all $n>0$, assume

$$\frac{4n-3}{4n+1}>\frac{5n-4}{5n+1}$$ $$(4n-3)(5n+1)>(4n+1)(5n-4)$$ $$20n^2-11n-3>20n^2-11n-4$$ $$-3>-4$$

which is true.

Answer 8

Begin by observing that $\frac{3996}{4000}=\frac{4995}{5000}$, and think of these fractions as $\frac{\mbox{wins}}{\mbox{games played}}$ for chess players $A$ ($3996$ wins) and $B$ ($4995$ wins). One additional win will do more to improve player $A$’s win percentage than it will player $B$’s.