Solution
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To solve the multiplication problem presented, we will follow the basic steps of arithmetic multiplication. The problem asks us to find the product of two numbers: 50 and 7.
Here is the step-by-step solution:
Step 1: Set up the multiplication
Here is the step-by-step solution:
Step 1: Set up the multiplication
Step 1: Set up the multiplication
Write down the two numbers you want to multiply, one below the other, aligning the digits by place value. In this case, since we are multiplying by a single-digit number, you can write the single-digit number (7) below the tens digit of the two-digit number (50).
```
50
× 7
```
Step 2: Multiply the digits
Multiply the single-digit number (7) by each digit of the two-digit number (50), starting from the rightmost digit. Since the rightmost digit of 50 is 0, multiplying 7 by 0 gives us 0.
```
50
× 7
----
0 (7 × 0 = 0)
```
Next, multiply the single-digit number (7) by the tens digit of the two-digit number (50), which is 5.
```
50
× 7
----
0 (7 × 0 = 0)
35 (7 × 5 = 35, but since we are multiplying with the tens place, it is actually 350)
```
Step 3: Add the results
Since we are multiplying by a single-digit number, there is no need to add multiple rows as you would in a larger multiplication problem. The result from the previous step is the final product.
```
50
× 7
----
350
```
Therefore, the product of 50 multiplied by 7 is 350.
In mathematical notation, this is expressed as:
This completes the multiplication problem.
Write down the two numbers you want to multiply, one below the other, aligning the digits by place value. In this case, since we are multiplying by a single-digit number, you can write the single-digit number (7) below the tens digit of the two-digit number (50).
```
50
× 7
```
Step 2: Multiply the digits
Multiply the single-digit number (7) by each digit of the two-digit number (50), starting from the rightmost digit. Since the rightmost digit of 50 is 0, multiplying 7 by 0 gives us 0.
```
50
× 7
----
0 (7 × 0 = 0)
```
Next, multiply the single-digit number (7) by the tens digit of the two-digit number (50), which is 5.
```
50
× 7
----
0 (7 × 0 = 0)
35 (7 × 5 = 35, but since we are multiplying with the tens place, it is actually 350)
```
Step 3: Add the results
Since we are multiplying by a single-digit number, there is no need to add multiple rows as you would in a larger multiplication problem. The result from the previous step is the final product.
```
50
× 7
----
350
```
Therefore, the product of 50 multiplied by 7 is 350.
In mathematical notation, this is expressed as:
This completes the multiplication problem.
Exercise 1-Prove that the geometric mean is less than or equal to the arithmetic mean.
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Exercise 1-Prove that the geometric mean is less than or equal to the arithmetic mean.
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This exercise is a preparation for the following exercise where we'll
prove that the geometric mean is less than or equal to the arithmetic mean.
Get us 2 parts.
In Part A, we have to show that if x and y satisfy x less than 1,
y bigger than 1,
then x plus y is bigger than x y plus 1.
This in turn will be used to prove Part B,
which later as I say,
it will be used to prove the AM-GM inequality in Part B,
we'll show that if a_1 to a_2,
are positive numbers whose product is 1,
then the sum is bigger or equal to n. Let's start with Part A. X is less than 1,
so 1 minus x is positive,
y is bigger than 1,
so y minus 1 is positive,
positive times positive is positive.
We get this. Now,
multiply it out we have this,
bring the minus 1 minus x, y
to the other side and we've proven this.
Let's get on to Part B,
which will do by induction on n. For each n,
we'll prove that for all a_1,
a_2 a_n which satisfy this, then this holds.
Now, if n equals 1,
then we only have 1 member,
it's a_1 and it's equal to 1 because the product of a_1 is just a_1 is 1.
So a_1 plus there's nothing to add,
so a_1 is bigger or equal to n, which is 1.
That's clear. If a_1 is 1,
then a_1 is bigger or equal to 1.
Let's do the induction part.
First of all, the induction hypothesis.
We'll fix some arbitrary n. For this n,
we know that if the product is 1,
then the sum is bigger or equal to n. From this will
show that it's true for n plus 1 that if we have a_1,
a_2 up to a_n plus 1,
whose product is 1,
then the sum will be bigger or equal to n plus 1.
This arrow means that this part is the given,
and this part is what we're going to show.
Divide into cases.
In the first case,
which is the uninteresting case,
all the A's are equal to 1.
Well, in this case,
if you add them all together,
1 plus 1 plus 1 plus 1,
but there's n plus 1 of them so we get n plus 1,
which is indeed bigger or equal to n plus 1.
Now case 2, we'll assume that at least 1 of the a_i is bigger than 1.
I claim that if 1 of the a_i is bigger than 1,
then another a_i well a_j is going to
be less than 1 because the product is going to be 1.
If you make 1 bigger, you got to make 1 smaller. Well, let's see.
Suppose on the contrary,
I mean proof by contradiction,
that for all the j from 1 to n plus 1,
then a_j is bigger or equal to 1.
That's the negation of less than 1.
In that case, if you look at a_1 up to a_n plus 1,
they're all bigger or equal to 1,
except that we know that a_i is strictly bigger than 1.
If you multiply all these together,
we get strictly bigger than 1 and that's a contradiction
because the product has to equal exactly 1.
We know that a_i is bigger than 1 and a_j is less than 1.
Now, we can certainly rearrange the order of the elements a_1 to a_n plus 1.
I mean, if we change the order of the product is still 1.
If we change the order here,
the sum doesn't change production,
the sum don't change if you rearrange the order.
We can push them to the end and assume that it's a_n
that's less than 1 and a_n plus 1 that's bigger than 1.
Now, bunched the last 2 together like 1 single element and
then all together we have a product of n elements which is 1.
Then we have that a_1 plus a_2 up to a_n,
a_n plus 1 as a single element,
this is bigger or equal to n. I'm going to do some trick to
somehow split this product to a sum.
That's where Part A will come in handy.
At x equals a_n and y equals a_n plus 1,
then x is less than 1 and y is bigger than 1.
We can use part a and get that x plus y is bigger than 1 plus x y.
If we rearrange that x y is less than x plus y minus 1,
this translates to a_n,
a_n plus 1 less than a_n plus n plus 1 minus 1.
If this is bigger or equal to n,
we replace this by something even bigger,
which is a_n plus n and plus or minus 1,
it'll be definitely bigger than n. We have the following.
Now, just bring the 1 over to the other side and drop
the brackets and we have that the sum is bigger or equal to n plus 1.
That's case 2.
Case 3 is almost the same as case 2.
Because if a_i is less than 1,
for some i,
I need to go back a moment.
The opposite of all of them equaling 1 means that at least 1 of them is not equal to 1.
We can break sub cases bigger than 1 and less than 1.
This is the other possibility that's left,
a_1 less than 1 for some i and exactly the same as in case 2.
We get that another a,
a_j is going to be bigger than 1.
If 1 is bigger than 1 is less than,
if 1 is less then 1 is bigger.
Then we can, just like before,
push them to the end and assume that it's a_n less than 1 and n plus 1 bigger than 1.
Once we have this,
then we can just continue exactly the same as in case 2.
There's nothing to do.
We've already done that.
That concludes this exercise.
This video explains how to prove the Arithmetic-Geometric Mean (AM-GM) Inequality, which states that the geometric mean of a set of positive numbers is less than or equal to the arithmetic mean. The proof is divided into two parts: Part A, which shows that if x and y satisfy x less than 1, y bigger than 1, then x plus y is bigger than x y plus 1; and Part B, which shows that if a_1 to a_n are positive numbers whose product is 1, then the sum is bigger or equal to n. Part A is proven by multiplying out the equation and rearranging the terms. Part B is proven by induction on n, with the induction hypothesis being that if the product of a_1 to a_n is 1, then the sum is bigger or equal to n. The proof is then divided into three cases: all a_i are equal to 1, at least one a_i is bigger than 1, and at least one a_i is less than 1. In each case, the proof is completed by rearranging the terms and using Part A.
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