SHOW THAT : 2^(2N) – 3N -1 is divisible by 9 FOR ALL N>=1
For every non-negative integer \( n \), the square root \( \sqrt{n} \) is either an integer or an irrational number. Proof (by contradiction): Suppose \( \sqrt{n} \) is rational but **not** an integer. Then there exist integers \( a \) and \( b \), with \( \gcd(a, b) = 1 \) and \( b \ne 1 \), such tRead more
For every non-negative integer \( n \), the square root \( \sqrt{n} \) is either an integer or an irrational number.
Proof (by contradiction):
Suppose \( \sqrt{n} \) is rational but **not** an integer.
Then there exist integers \( a \) and \( b \), with \( \gcd(a, b) = 1 \) and \( b \ne 1 \), such that:
\[
\sqrt{n} = \frac{a}{b}
\]
Squaring both sides:
\[
n = \left( \frac{a}{b} \right)^2 = \frac{a^2}{b^2}
\Rightarrow a^2 = n b^2
\]
This implies that \( b^2 \) divides \( a^2 \). But since \( \gcd(a, b) = 1 \), it follows that \( \gcd(a^2, b^2) = 1 \) as well. Hence, the only way \( b^2 \mid a^2 \) can be true is if \( b^2 = 1 \), which means:
\[
b = 1
\Rightarrow \sqrt{n} = \frac{a}{1} = a \in \mathbb{Z}
\]
This contradicts our assumption that \( \sqrt{n} \) is rational **but not** an integer.
Conclusion:
If \( \sqrt{n} \) is rational, then it must be an integer.
Therefore, if \( \sqrt{n} \) is not an integer, it must be irrational.
\[
\boxed{\text{For all } n \in \mathbb{N}_0,\ \sqrt{n} \in \mathbb{Z} \cup (\mathbb{R} \setminus \mathbb{Q})}
\]
Proof by Mathematical Induction: Base Case: Let \( n = 1 \) \[ 2^{2 \cdot 1} - 3 \cdot 1 - 1 = 2^2 - 3 - 1 = 4 - 3 - 1 = 0 \] \[ \Rightarrow 9 \mid 0 \quad \text{(True)} \] Inductive Hypothesis: Assume that for some \( n = k \), the expression is divisible by 9: \[ 2^{2k} - 3k - 1 \equiv 0 \pmod{9}Read more
Proof by Mathematical Induction:
Base Case: Let \( n = 1 \)
\[
2^{2 \cdot 1} – 3 \cdot 1 – 1 = 2^2 – 3 – 1 = 4 – 3 – 1 = 0
\]
\[
\Rightarrow 9 \mid 0 \quad \text{(True)}
\]
Inductive Hypothesis: Assume that for some \( n = k \), the expression is divisible by 9:
\[
2^{2k} – 3k – 1 \equiv 0 \pmod{9}
\]
That is,
\[
2^{2k} – 3k – 1 = 9m \quad \text{for some integer } m
\]
To Prove: \( 2^{2(k+1)} – 3(k+1) – 1 \) is also divisible by 9.
LHS:
\[
2^{2(k+1)} – 3(k+1) – 1 = 2^{2k + 2} – 3k – 3 – 1
\]
\[
= 4 \cdot 2^{2k} – 3k – 4
\]
Now subtract the inductive hypothesis:
\[
[4 \cdot 2^{2k} – 3k – 4] – [2^{2k} – 3k – 1] = 3 \cdot 2^{2k} – 3
\]
\[
= 3(2^{2k} – 1)
\]
Now, observe that \( 2^{2k} \equiv 1 \pmod{3} \), so \( 2^{2k} – 1 \equiv 0 \pmod{3} \)
Hence, \( 3(2^{2k} – 1) \equiv 0 \pmod{9} \)
Therefore, the expression for \( n = k+1 \) is also divisible by 9.
Conclusion:
\[
See less\boxed{2^{2n} – 3n – 1 \text{ is divisible by } 9 \text{ for all } n \geq 1}
\]