Question

Which is greater?
$\rm \large 88^{89} \: \: or \: \: 89^{88}$
​

Answer 1

Answer:

both are equal

Step-by-step explanation:

88

×89=7832.89×88=7832

Answer 2

$\Large\text{Topic: [Mathematical induction]}$

Let $\largen=88$. Then, -

$\cdots\longrightarrow88^{89}=n^{n+1},\ 89^{88}=(n+1)^{n}.$

$\Large\text{[Assertion]}$

$\large\text{ \cdots\longrightarrow\boxed{n^{1+\frac{1}{n}}>n+1\text{ for any integers }n\geq3.} }$

To prove this, we use mathematical induction.

$\Large\text{[Base case]}$

The base case for $\largen=3$.

To prove: $\large\text{ 3^{1+\frac{1}{3}}>3+1. }$

Dividing by $\large3$, -

$\large\text{ \cdots\longrightarrow3^{\frac{1}{3}}>1+\dfrac{1}{3}. }$

Raising both numbers by 3, -

$\cdots\longrightarrow3>\left(\dfrac{4}{3}\right)^{3}$

$\cdots\longrightarrow3>\dfrac{64}{27}.\ \text{[True]}$

This inequation is true. The base case is confirmed.

$\Large\text{[Mathematical induction]}$

If the inequality is true, we would get -

$\text{ \cdots\longrightarrow k^{1+\frac{1}{k}}<k+1\text{ for any integers }k\geq3. }$

Let's put $\largen=k+1$ to prove for all numbers above the base case.

$\text{ \cdots\longrightarrow k^{1+\frac{1}{k+1}}<k+2. }$

From the exponents, we can compare numbers.

$\large\text{ \cdots\longrightarrow\boxed{k^{1+\frac{1}{k}}>k^{1+\frac{1}{k+1}}\text{ for any natural numbers }k} }$

From this fact, we can derive an inequality.

$\text{ \cdots\longrightarrow k^{1+\frac{1}{k+1}}<k^{1+\frac{1}{k}}<k+1<k+2 }$

$\text{ \cdots\longrightarrow k^{1+\frac{1}{k+1}}<k+2\ \text{[True]} }$

So, by mathematical induction, the given inequality is true for natural numbers $\largen\geq3$ $\large\blacksquare$

$\Large\text{[Theorem]}$

$\large\text{ \cdots\longrightarrow\boxed{n^{1+\frac{1}{n}}>n+1\text{ for any integers }n\geq3.} }$

Raising $\largen$ on both sides, -

$\large\text{ \cdots\longrightarrow\boxed{n^{n+1}>(n+1)^{n}\text{ for any integers }n\geq3.} }$

Hence, we can say that $\large88^{89}>89^{88}.$

$\Large\text{[Final answer]}$

Hence, -

$\large\text{ \cdots\longrightarrow\boxed{88^{89}>89^{88}.} }$