GO Classes CS/DA 2025 | Weekly Quiz 2 | Fundamental Course | Question: 3

Answer 1

Detailed Video Solution - Weekly Quiz 2

Annotated Notes - Weekly Quiz 2 Solutions

$\text{S1}:$ Suppose $a, b \in \mathbb{Z}$. If $a^2\left(b^2-2 b\right)$ is odd, then both $a$ and $b$ are odd.
(Suggestion: Try contrapositive proof.)
Proof. (Contrapositive) Suppose it is not the case that $a$ and $b$ are odd. Then, by DeMorgan's Law, $a$ is even or $b$ is even. Let us look at these cases separately.

Case 1. Suppose $a$ is even. Then $a=2 c$ for some integer $c$.
Thus $a^2\left(b^2-2 b\right)=(2 c)^2\left(b^2-2 b\right)=2\left(2 c^2\left(b^2-2 b\right)\right)$, which is even.

Case 2. Suppose $b$ is even. Then $b=2 c$ for some integer $c$.
Thus $a^2\left(b^2-2 b\right)=a^2\left((2 c)^2-2(2 c)\right)=2\left(a^2\left(2 c^2-2 c\right)\right)$, which is even.
Thus in either case $a^2\left(b^2-2 b\right)$ is even, so it is not odd.
(NOTE: A third case where both $a$ and $b$ are even is not necessary.
In that case $a$ is even, a scenario addressed in Case 1.)

$\text{S2}:$ Suppose $a, b \in \mathbb{Z}$. If $25 \nmid a b$, then $5 \nmid a$ or $5 \nmid b$.
Proof. (Contrapositive) Suppose it is not the case that $5 \nmid a$ or $5 \nmid b$. Then (using DeMorgan's Law), we have $5 \mid a$ and $5 \mid b$.
By definition of divisibility, this gives $a=5 k$ and $b=5 \ell$ for some $k, \ell \in \mathbb{Z}$.
Then $a b=(5 k)(5 \ell)=25 k \ell$, that is, $a b=25 c$ for $c=k \ell \in \mathbb{Z}$.
Therefore $25 \mid a b$, by definition of divisibility.
Thus it is not the case that $25 \nmid a b$.

Comments

In this question for S2 statement:

if 25 does not divides ab means two cases:

1. a=5k1, b=k2  where k2 is not multiple of 5
2. a=k1, b=k2  where k1 and k2 are not multiple of 5

for 1^st condition 5 does divide a and 5 does not divide b.

for 2^nd condition 5 does not divide both a and b.

from the statement its clear that  5 should not divide a OR 5 should not divide b.

Hence statement is true because of OR . Both conditions can be satisfied or either one must be true.

Here i lost marks due to misinterpretation of OR word.

Don't make same mistake as me :(

Answer 2

Direct Proof for S1:

Given $(a^2)(b^2-2b)$ is odd where a,b $\in \mathbb{Z}$

To Prove –> a and b are both odd.

We know, $odd*odd = odd$

and we have $(a^2)*(b^2-2b) = odd.$

$\therefore a^2 = odd$ and $(b^2-2b) = odd$
     $a^2 = odd \Rightarrow a = odd.$

Also we know,   even – odd $\|$ odd – even = odd.
Since, $(b^2-2b) = odd $ and $2*b = even$
    $\Rightarrow b^2 = odd$
    $\Rightarrow b = odd.$

$\therefore$ a and b are both odd.

Hence Proved.

Comments

 

.