GO Classes CS/DA 2025 | Weekly Quiz 2 | Fundamental Course | Question: 2

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GO Classes CS/DA 2025 | Weekly Quiz 2 | Fundamental Course | Question: 2

GO Classes asked Mar 8, 2023 • edited Mar 9 by shadymademe

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commented Mar 9, 2023

If you know these 2 results
1) if $a$ is odd then $a^{2}$ is odd and viceversa
2) if $b$ is odd then $b^{2}$ is odd and viceversa

then the given statements can be easily proven

S1: If a, b are integers and $a+b$ is even, then $a^2+b^2$ is even
Proof: $a+b$ is even means both a, b are even or a,b are odd
Case 1: a, b both are even => $a^{2}$ and $b^{2}$ both are even, hence $a^2+b^2$ is even (even + even is even)
Case 2: a, b both are odd => $a^{2}$ and $b^{2}$ both are odd, hence $a^2+b^2$ is even (odd + odd is even)

S2: If a, b are integers and $a+b$ is odd, then $a^2+b^2$ is odd
Proof: Since $a+b$ is odd. either exactly one of a or b is odd (not both) => either exactly one of $a^{2}$ or $b^{2}$ is odd, hence $a^2+b^2$ is odd (odd+even is odd)

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If you know these 2 results
1) if $a$ is odd then $a^{2}$ is odd and viceversa
2) if $b$ is odd then $b^{2}$ is odd and viceversa

then the given statements can be easily proven

S1: If a, b are integers and $a+b$ is even, then $a^2+b^2$ is even
Proof: $a+b$ is even means both a, b are even or a,b are odd
Case 1: a, b both are even => $a^{2}$ and $b^{2}$ both are even, hence $a^2+b^2$ is even (even + even is even)
Case 2: a, b both are odd => $a^{2}$ and $b^{2}$ both are odd, hence $a^2+b^2$ is even (odd + odd is even)

S2: If a, b are integers and $a+b$ is odd, then $a^2+b^2$ is odd
Proof: Since $a+b$ is odd. either exactly one of a or b is odd (not both) => either exactly one of $a^{2}$ or $b^{2}$ is odd, hence $a^2+b^2$ is odd (odd+even is odd) — Vineeth Rambhiya, Mar 9, 2023

GO Classes · Answer 1 · 2023-03-08T02:38:49+0000

Detailed Video Solution - Weekly Quiz 2

Annotated Notes - Weekly Quiz 2 Solutions

$\text{S1}:$
Prove: If $a, b \in \mathbb{Z}$ and $a+b$ is even, then $a^2+b^2$ is even.
(Hint: Try direct.)

Proof (Direct) Suppose that $a+b$ is even.
This means $a+b=2 k$, for some integer $k$.
Then $(a+b)^2=(2 k)^2$, and this becomes $a^2+2 a b+b^2=4 k^2$.
Transposing the above equation, we obtain $a^2+b^2=4 k^2-2 a b=2\left(2 k^2-a b\right)$.
Therefore, we have $a^2+b^2=2(c)$, where $c=2 k^2-a b \in \mathbb{Z}$.
Thus, by the definition of an even integer, $a^2+b^2$ is even.

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GO Classes CS/DA 2025 | Weekly Quiz 2 | Fundamental Course | Question: 2

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