Help in understanding the logic expressions

Question

i have exercise do not understand the operation marked in red. How do I make the first expression marked the second expression red Squared marked in red? I did not understand how it is the same and how do I get it?

Answer 1

The triple line means "congruent", and saying $$a\equiv b\mod c$$

simply means that $a$ and $b$ both have the same remainder when divided by $c$.

In your case, $3n^3 + 9n^2 + 15n + 9$ has the same remainder as $3n^3 + 6n$ when divided by $9$

For more detail, check http://mathworld.wolfram.com/Congruence.html

---

An important thing to know is that

$a\equiv b\mod c\iff a-b$ is divisible by $c$

This is easy to see, since we can always write $a=k_a\cdot c + n_a$ and $b=k_b\cdot c + n_b$, where $0\leq n_a,n_b

• If $a-b$ is divisible by $c$, then $a-b=(k_a-k_b)\cdot c + (n_a-n_b)$ is divisible by $c$. Since $(k_a-k_b)c$ is divisible by $c$, so is $n_a-n_b$, and since $-c < n_a-n_b < c$, this is only possible if $n_a-n_b=0$, so $a\equiv b\mod c$.
• If $a\equiv b\mod c$, then $n_a=n_b$, which means that $a-b=(k_a-k_b)c$ and $a-b$ is divisible by $c$.

Having this what you need for your claim is the following claim:

For any integer $a$ and any integer $k$, $a + k\cdot b\equiv a\mod b$

The claim is true because $(a+k\cdot b)- a = k\cdot b$ which is divisible by $b$.

Now, if you write $3n^3 + 9n^2 + 15n + 9$ as $$9n^2 + 9n + 9 + 3n^3 + 6n$$ everything should be clear.

Answer 2

We have $3n^3 + 9n^2 + 15n + 9$.

To take it simple we have one term $9n^2$.

And $9n^2\mod 9 = 0$. As term is multiple of 9.

Then we have next term $15n$

$15n = 9n + 6n \mod 9 = 6n$. As one term is multiple of 9.

Last term is $9$

So $9 \mod 9 = 0. As divisible by 9.

Combining all these,

$3n^3 + 9n^2 + 15n + 9 = 3n^3 + 6n $

Hope clears to you now.