Hi. I'm Vladimir Podolskii.

Today we are going to discuss division by two.

So we have integer numbers.

Let's consider division of these numbers by two.

There are two possible remainders that numbers can have when we divided them by two.

The remainder can be zero or it can be one.

If a remainder of some number is zero,

then this number is divisible by two,

and these numbers are called even numbers.

If the remainder is one,

then this number is not divisible by two and these numbers are called odd numbers.

Okay. Let's consider the following problem.

Suppose we have two classes.

One, we have A students,

and one, we have B students.

We unite these classes.

We would like to now to split all of the students into pairs to work on a project.

Is it possible to do this,

to split all students into pairs if A is even and B is odd?

The same question, we can ask if both of the numbers are even,

and the same question we can ask if A and B are both odd.

Okay. Let's start with the first question.

Suppose A is even and B is odd, what we can do now?

Now, we can split all students into pairs in the first class because A is even,

so we can split all students into pair.

A is divisible by two.

In the second class,

we can split all students into pairs,

all students except one into pairs.

There are odd number of students in B so the remainder will be one,

so we can split everyone into pairs,

all students into pairs, except one.

One student will be left.

So, overall, we have one student left and the answer is no,

this is not possible.

Okay. Now, consider the next case.

If both A and B are even,

then we can just split students into pairs in both classes

separately since there are even number of students in the first class,

we can split them into pairs there,

and since there are even number of students in the second class,

then, again, we can split them into pairs.

So, here the answer is yes, we can do it.

We can just split students into pairs in each class separately.

Okay.

Now, the last question.

If both A and B are odd,

then what we can do,

we can split students into pairs in the first class,

and one student will be left.

A is odd and we can split students in the second class,

and one student will be left again.

B is also odd.

So we have splitted almost all students,

but two students are left,

one from the first class and one from the second class.

So now we can pair our two remaining students and everyone will have a pair.

So we have split our students into pairs in all classes.

So, this is also possible and the answer is yes here again.

Okay, now, let's restate this in our mathematical terms.

So, suppose we have two numbers, A and B

and we know the remainders of A and B when we divided them by 2.

Can we deduce the remainder of a+b when we divided by two?

Okay, it turns out that we can.

The integers was even as in the previous problem.

Okay, let's consider cases if both A and B are

even so the remainders are zero both for A and B,

then we have the form 2 x q1 and 2 x q2.

And so, the sum has a form 2 x (q1 + q2).

It is an even number.

The remainder here is also zero.

So, if both A and B are zero,

so the sum is also if both A and B has the remainder is zero,

then the sum also the remainder zero.

Now, if one of the numbers is even and the other is odd,

then the even number has a form 2 x q1,

and the odd number has a form 2 x q2 + 1.

Now, the sum has a full one for 2 x q1 + q2 + 1.

And this is an odd number.

So the remainder here is one.

So if one of A and B is even and the other is odd, then the sum is odd.

In the final case is the full one,

suppose both A and B are odd.

And then both all of them has the form 2 x q1 + 1 and 2 x q2 + 1.

So the sum will be called to 2 x(q1 + q2)+ 1.

Clearly, this number is divisible by two,

so it is even.

Overall, we have the following table for the remainders of a+b when we divide them by 2.

The rules of the table correspond to the remainders of A,

and the columns of the table correspond to the remainders or B.

And then in the cell of the table,

we write the remainder of a+b.

So if we consider the second row and the second column,

then we consider the case when the remainder of

A is one and the remainder of B is also one.

And so, the remainder of a+b as we establish is zero.

a+b is even if both A and B are odd.

Now, consider the following situation.

Supposed, we know the remainder of A when we divide it by two.

Can we deduce from it?

What is the remainder of -a?

-a is also an integer number,

so it also has its remainder.

Can we say what is the remainder of -a?

Yes. Yes, we can.

And it is exactly the same as remainder of A.

So why it is so?

Note that if A is divisible by two then -a is also divisible by two, and vice versa.

If -a is divisible by two, then A is divisible by two.

So, if A has the remainder zero,

then -a has the remainder zero.

If A has the remainder 1,

it is not divisible by two,

then minus a is not divisible by two as well,

and the remainder of -2 is also 1.

Now, we can deduce the same result we obtained for- For the sum of the numbers,

we can do the same for the difference of two numbers.

If we consider the difference of two numbers A and B,

if we consider the remainder of this number when we divide it by

two when this remainder is the same as the remainder of a+b.

And the reason is that -b and +b has the same remainder,

so instead of subtracting B,

we can add -B which is the same as to add B,

so the remainder is the same.

So for subtraction, we have the same table as for addition.

So, again, rules could respond to

the remainder of A columns correspond to remainders of B,

and then the cells we write the remainders of a - b.

It is the same as for addition.

Now, we can consider a product.

Suppose we know the remainder of two numbers A and B,

and when we divide them by two.

Can we tell something about the remainder of their product a x b?

And it turns out, again, we can do it.

The remainder of a + a x b is

completed in their mind by the remainders of A and the remainders of B.

And let's consider the cases.

And there are two different cases here.

If at least one of the numbers is even,

so it is divisible by two,

and then the product is also divisible by two,

note that if one of the numbers we are multiplying is divisible by two,

then the product is also divisible by two.

So, if the remainder of at least one of the numbers is zero,

then the remainder of the product is also zero.

And the second case is when both A and B are odd,

so then they have a form 2 x q1 + 1 and 2 x q2 + 1.

So they are not divisible by two,

so the remainder is 1.

And then let's just write down their product.

It is q1 x 2 + 1 and q2 x 2 + 1.

And then we can open the brackets and see

that almost all sum will have multiplayer two.

And so overall we will have q1 x q2 + q1 + q2.

And all of this is multiplied by 2 + 1.

And this is an odd number, so it's easy to see that

the last expression is exactly the expression for division of the remainder.

And the remainder here is 1,

so the number is odd.

And so, we have the following table for the remainders of a x b when we divide it by two.

Again, rows here correspond to the remainders of A,

columns correspond to reminders of B.

And in the cell we write the remainders of a x b.

So here, the remainder of a x b is one if

and only if the remainders of both A and B are 1.

In all other cases,

the remainder of a x b is zero.

Now, we can apply these rules to some specific problem.

Suppose, we have the following expression,

so some numbers and we apply addition,

multiplication to them and subtraction,

and we would like to know what is

the remainder of the result and number when we divide them by two.

Is this number divisible by two or not?

Is it even or odd?

Of course, we can do the calculation and then apply divisibility test for two.

We have to look at the last digit.

But, actually, using our rules they can do it.

In a more simple way we can do without any calculations.

So everything is simple.

[inaudible]. Note that the remainders of the results of operations addition,

subtraction, multiplication depend only on

the remainders of numbers we apply our operation to.

We can just employ our operations directly to remainders,

and then we will obtain the remainder of this number when we divide it by two.

So we can just substitute all numbers by their remainders.

So note that 374 and 48 are even,

so the remainders are zero, and 419, 467,

625 are odd, so their remainders are one.

So, we obtain this expression.

And now we can just do a calculation according to our tables,

1 x 0 for the remainder is zero.

Now we have 1 + 0 in the brackets,

so the remainder will be one.

If we add odd number and even number, the result is odd.

Now, we multiply zero by the bracket,

and the result is always zero.

So we have 0 - 1.

And for the remainder, it means that the result is 1.

So, finally, we have the result.

The remainder of the result in expression is one.

And this is the answer to our problems,

so we can find the answer to the problem now without calculations.

So the resultant remainder is 1.

And so the remainder of our initial number when we divide it by 2 is 1.

And so, we can compute the remainder here without complicated computations.

We do not have to compute the number

to tell what is the remainder of this number when we divide it by two.