GO Classes 2025 | Weekly Quiz 2 | Fundamental Course | Question: 1

Answer 1

Detailed Video Solution - Weekly Quiz 2

Annotated Notes - Weekly Quiz 2 Solutions

$\text{S2}:$
Prove: Suppose $a \in \mathbb{Z}$. If $a^3$ is not divisible by $8,$ then $a$ is odd.
(Hint: Try contrapositive.)
 

Proof (Contrapositive) Suppose that $a$ is not odd, that is, that $a$ is even.
Then $a=2 c$, for some integer $c$, by definition of an even integer.
Consequently, $a^3=(2 c)^3=8 c^3$.
Therefore $a^3=8 d$ for the integer $d=c^3$.
By definition of divides, this implies that $8 \mid a^3,$ so it is not true that $a^3$ is not divisible by $8.$

$\text{S1}:$ Hint: Try direct

Comments

Adding the proof of S1:

$a^{2}\mid b$ then, $b = n_{1}a^{2}$     …(i)

$b^{3}\mid c$ then, $c = n_{2}b^{3}$      …(ii)

From (i) and (ii) we have:

$c = n_{2}(n_{1}(a^{2}))^{3}$ =  $ (n_{1}^{3}n_{2})a^{6}$

$\therefore$  $a^{6}\mid c$   proved

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https://gateoverflow.in/?qa=blob&qa_blobid=12927433878391882559why cant we do like this ...going through this method answer is not coming

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@Maverick09, your method (direct method) tells you that $a$ is either odd or even. So, it can't reach the actual conclusion that $a$ is odd. 

Contrapositive method tells you that $a$ is odd. 

Note that direct method didn't give you wrong answer... But it didn't give you any answer at all. 

It is like asking "Is John male?". Person A (direct method in this analogy) says "John is either male or female".. Now this answer is Not wrong, But it is not giving you actual information either. Person B (contrapositive proof in this analogy) says "Yes, John is male". 

Answer 2

OPTION C SEEM TO BE TRUE :)

Comments

A little bit of mistake while writing the final answer of 1) i.e. $a^{6}$ | c

Bro you’ve done everything right , just made a small mistake in the final line…

$a^{6}\times k^{3}\times k_{1} = c \Rightarrow a^{6}\mid c$

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Thnks I will edit for sure ...

Answer 3

Easy and Simple just Look once

Answer 4

$S_1 :$

$a^2|b$ $\text{means } a^2 \text{ divides } b \text{, which means}$ $b=a^2k_1\text{, where } k_1\in \mathbb Z \qquad \ldots (1)$

$\text{Similarly, for } b^3|c \text{ , } c=b^3k_2  \text{ , where } k_2\in \mathbb Z \qquad \ldots (2)$

$\text{Substituting the value of b from equation 1 to 2 we get:}$

$c=(a^2k_1)^3k_2=a^6(k_1k_2) $

$\therefore a^6|c$

 

$S_2 :$

$\text{This can be proved by proof by contraposition. It makes use of the fact that } \mathbf{P} \to \mathbf{Q} \text{ is equivalent to } \neg \mathbf{Q} \to \neg \mathbf{P}$

$\mathbf{P} : a^3 \text{ is not divisible by } 8$

$\mathbf{Q} : a \text{ is odd}$

$\text{So, we have to prove, if a is even }(\neg \mathbf{Q}) \text{, then }a^3 \text{ is divisible by  }8 (\neg \mathbf{P})$

$\text{Let’s say } a \text{ is even, then:}$

$a=2b, \text{ for some integer }b\text{ , by definition of an even integer}$

$\text{Then, } a^3=(2b)^3=8b^3$

$\text{From this we can clearly see that } 8|a^3.$

 

$\text{So, }S_1 \text{ and, } S_2 \text{ are both true. } $