$∀ ∈ ℕ, m^2 + 3$ is even ⟶ $m$ is odd
$P = m^2 + 3$ is even
$Q = m$ is odd
By contrapositive: $¬Q ⟶ ¬P$
- $¬Q m$ is even, $m = 2k$
- $¬P m^2 + 3$ is odd $= (2k)^2 + 3$
Help what I should do now after this step!
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2Do you realize that $(2k)^2+3$ is odd for every $k$? – drhab Oct 05 '23 at 14:45
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No , how that ! – Amanda Oct 05 '23 at 15:54
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1$(2k)^2+3=4k^2+3=2(2k^2+1)+1$ – drhab Oct 05 '23 at 16:33
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Can you explain how convert the number 3 to 1 because i think it gives a fractional number? – Amanda Oct 05 '23 at 17:35
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1$4k^2+3=(4k^2+2)+1=2(2k^2+1)+1$. I run out of ways to explain further. The sum of an odd and an even integer is always odd. – drhab Oct 05 '23 at 17:42
1 Answers
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As stated in the comments, a number is odd if and only if it is of the form $2a+1$ for some integer $a$.
$(2k)^2 + 3 = 4k^2 + 3 = 4k^2 + (2 + 1) = (4k^2 + 2) + 1 = 2(2k^2 + 1) + 1$.
Since $k$ is an integer, $k^2$ is an integer, so $2k^2+1$ is an integer. Let $a=2k^2+1$. Then $m=2a+1$, an odd integer.
IAAW
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