Maths: Which One Is Bigger, 1000^1001 or 1001^1000?

The question is as such:

$1000^1001$❓$1001^1000$

⬆️ We want to find the❓operator.

Well, not... "we" per se. You might think — WHY am I involved in this? 🤔 I'm not obliged to solve that, am I? — Indeed. Do excuse me.

Let me rephrase that.

I want to find the ❓ operator.

Mwahahaha.

It's the reasonind behing this post. 🦹

Right. So.

I scribbled that using logarithmic. Here it goes.

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The Scribble

$1000^1001❓1001^1000$

Let me assume:

$1000 = a$

Therefore:

$1001 = (a + 1)$

So this:

$1000^1001❓1001^1000$

Becomes:

$a^(a+1)❓(a + 1)^a$

➡️ I then place log (logarithmic function) on both sides.

This form:

$a^(a+1)❓(a + 1)^a$

➡️ Becomes:

$log_a$$(a^(a+1))$ $❓$ $log_a$$((a + 1)^a)$

Continue by implementing —

logarithmic properties.

The form above can be transformed into this:

$(a + 1) log_a(a)❓(a) log_a(a + 1)$

I can simplify $log_a(a) = 1$. Thus:

$(a + 1) (1)❓(a) log_a(a + 1)$

$log_a(a + 1) > 1$ — very near 1 because $a = 1000$.

Therefore, I can safely simplify that —

$log_a(a + 1) = 1$

Final simplification:

$(a + 1) (1)❓(a) (1)$

$(a + 1)❓(a)$

Anyway —

💡

For any number, adding one gives a result larger than the number itself — naturally.

Thus:

$(a + 1) > a$

Meaning:

$1000^1001$ is greater than $1001^1000$

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$1000^1001$ $>$ $1001^1000$

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The answer is $>$

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Pattern

As I've observed, the pattern starts from any integer greater than $2$.

Let me try that.

$1$ and $2$

$1^2$❓$2^1$
$1$❓$2$
$1 < 2$

$2$ and $3$

$2^3$❓$3^2$
$8$❓$9$
$8 < 9$

$3$ and $4$

$3^4$❓$4^3$
$81$❓$64$
$81$ $>$ $64$

$4$ and $5$

$4^5$❓$5^4$
$1024$❓$625$
$1024$ $>$ $625$

$5$ and $6$

$5^6$❓$6^5$
$15625$❓$7776$
$15625$ $>$ 7776

$6$ and $7$

$6^7$❓$7^6$
$279936$❓$117649$
$279936$ $>$ 117649

$7$ and $8$

$7^8$❓$8^7$
$5764801$❓$2097152$
$5764801$ $>$ 2097152

•••

And so forth.

Ah. The pattern.

Once $a$ gets bigger than 2, $a^{(a+1)}$ will always comfortably beat $(a+1)^a$ every single time.

Establishing that —

$a^{(a+1)} > (a+1)^a$

for $a > 2$.

Naturally.

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