Cube of n minus 23 Greater than Square of (4n-7)
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Theorem
Let $n \in \Z$ such that $n \ge 12$.
Then:
- $n^3 - 23 > \paren {4 n - 7}^2$
Proof
The proof proceeds by induction.
For all $n \in \Z$ such that $n \ge 12$, let $\map P n$ be the proposition:
- $n^3 - 23 > \paren {4 n - 7}^2$
Basis for the Induction
$\map P {12}$ is the case:
\(\ds 12^3 - 23\) | \(=\) | \(\ds 1728 - 23\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1705\) | ||||||||||||
\(\ds \paren {4 \times 12 - 7}^2\) | \(=\) | \(\ds 41^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1681\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 12^3 - 23\) | \(>\) | \(\ds \paren {4 n - 7}^2\) |
Thus $\map P {12}$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 12$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $k^3 - 23 > \paren {4 k - 7}^2$
from which it is to be shown that:
- $\paren {k + 1}^3 - 23 > \paren {4 \paren {k + 1} - 7}^2$
Induction Step
This is the induction step:
\(\ds \paren {k + 1}^3 - 23\) | \(=\) | \(\ds k^3 + 3 k^2 + 3 k + 1 - 23\) | ||||||||||||
\(\text {(1)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {k^3 - 23} + \paren {3 k^2 + 3 k + 1}\) | |||||||||||
\(\ds \paren {4 \paren {k + 1} - 7}^2\) | \(=\) | \(\ds \paren {\paren {4 k - 7} + 4}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {4 k - 7}^2 + 2 \times 4 \times \paren {4 k - 7} + 4^2\) | Square of Sum | |||||||||||
\(\text {(2)}: \quad\) | \(\ds \) | \(=\) | \(\ds \paren {4 k - 7}^2 + 32 k - 40\) | simplifying |
Then we calculate:
\(\ds \paren {3 k^2 + 3 k + 1} - \paren {32 k - 40}\) | \(=\) | \(\ds 3 k^2 - 29 k + 40\) | ||||||||||||
\(\ds \) | \(\ge\) | \(\ds \paren {3 \times 12} k - 29 k + 40\) | as $k \ge 12$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 7 k + 40\) | simplification | |||||||||||
\(\ds \) | \(>\) | \(\ds 0\) | ||||||||||||
\(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \paren {3 k^2 + 3 k + 1}\) | \(>\) | \(\ds \paren {32 k - 40}\) |
Thus we have:
\(\ds \paren {k^3 - 23}\) | \(>\) | \(\ds \paren {4 k - 7}^2\) | Induction Hypothesis | |||||||||||
\(\ds \paren {3 k^2 + 3 k + 1}\) | \(>\) | \(\ds \paren {32 k - 40}\) | from $(3)$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {k^3 - 23} + \paren {3 k^2 + 3 k + 1}\) | \(>\) | \(\ds \paren {4 k - 7}^2 + \paren {32 k - 40}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {k + 1}^3 - 23\) | \(>\) | \(\ds \paren {4 \paren {k + 1} - 7}^2\) | from $(1)$ and $(2)$ |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{\ge 12}: n^3 - 23 > \paren {4 n - 7}^2$
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): induction: 1. (in mathematics)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): induction: 1. (in mathematics)