Food for thought

## Solution part 1

Two fractions are shown below.

$\textrm{Fraction} \ A: \frac{10^a +1}{10^a +2} \qquad \textrm{Fraction} \ B: \frac{10^b + 1}{10^b + 2}$

For what values of $a$ and $b$ will fraction A be greater than fraction B?

In order to compare two fractions, they need to have the same denominator. In this case we could multiply the two denominators together to give a common denominator of $(10^a + 2)(10^b +2)$.

Writing each fraction with this common denominator we get $\textrm{Fraction} \ A: \frac{10^a +1}{10^a + 2}=\frac{(10^a + 1)(10^b + 2)}{(10^a+2)(10^b + 2)}$ and $\textrm{Fraction} \ B: \frac{10^b + 1}{10^b + 2}=\frac{(10^b+1)(10^a+2)}{(10^b+2)(10^a+2)}.$

Is this common denominator, $(10^a+2)(10^b+2)$, a positive or negative quantity?

• Does it depend on the values of $a$ and $b$?
• Does it make a difference to our next steps when we come to compare the two fractions?

We can now compare $\textrm{Numerator} \ A: (10^a + 1)(10^b + 2) \quad \textrm{and} \quad \textrm{Numerator} \ B: (10^b +1)(10^a+2).$

There are several ways in which we could do this. One strategy might be to expand the brackets and compare the terms $\textrm{Numerator} \ A: (10^a+1)(10^b+2)=10^{a+b}+2\times10^a + 10^b + 2$ and $\textrm{Numerator} \ B: (10^b+1)(10^a+2)=10^{a+b}+2\times10^b+10^a+2.$

Only the middle terms differ so if we consider this part of the numerator of $A$: $C: 2\times10^a+10^b$ and the equivalent part of the numerator of $B$: $D: 2\times10^b+10^a.$

If $a>b$ then in expression C the first term will dominate and in expression D the second term will dominate. Since the first term of C is double the second term in D, expression C must be greater.

Similarly, if $b>a$ then in expression C the second term will dominate and in expression D the first term will dominate. Since the first term of D is double the second term of C expression D must be greater.

This means that overall if $a>b$ then $\textrm{Fraction} \ A > \textrm{Fraction} \ B$ and if $b>a$ then $\textrm{Fraction} \ B > \textrm{Fraction} \ A.$

We could have reached this solution more quickly if we had noticed at the above expressions (showing the differing middle terms) can be reduced to $C: 10^a \quad D: 10^b.$

Thinking about this graphically we see that $10^x$ is an increasing function and so fraction A will be greater than fraction B when $a>b$.

We should check that this makes sense when we substitute values in place of $a$ and $b$. Substituting $a=1$ and $b=0$ into fractions A and B gives $A: \frac{10^1+1}{10^1+2} = \frac{11}{12}$ and $B: \frac{10^0+1}{10^0+2} = \frac{2}{3}.$

Then we can show that $\frac{11}{12}>\frac{2}{3}$ and our solution holds.