Let's say that we weren't dealing with reals.

Write 5 and 1 4 as an improper fraction.

Let's say I had twenty three over five.

3 left over, but it's 3 of your 4, or 3 4 left over.

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These two things are equivalent.

Sorry, I don't like mixed numbers.

If you divide 4 by 2, you get 2, so this is equal to 3 2.

So 1 and 7 68.

It's that simple.

There are also no local variable declarations.

We make a variable, greeting, initialize to hola . Variable greeting is hola .

So we could go 1, 2, and then 1 5.

Write 5.1 4 as an improper fraction

And then what's the derivative of y squared with respect to x?

I'll assume it's just a function of x.

So 15 is greater than or equal to 8, so we are dealing with an improper fraction.

Let me just do a little side example here.

So now, let's see this algorithm in action.

It's some place halfway between two and three.

So let's go the other way.

Well an improper fraction is a fraction where the numerator is larger than the denominator.

Identify 15 eights, or 15 over 8, as a proper or an improper fraction. or an improper fraction.

Now, there's two ways to do this.

It's 63, over 1 times 5.

Write 7 4 as a mixed number.

And so this will become 6 over 4.

Actually, let's do it that way.

It was hot. It wasn't hot enough.

4 and 1 8.

So let's start with the latter.

Or, we could reverse that causal flow.

So this can be rewritten as the principal square root of negative 1 times 52.

So, gradient descent so far seems to be doing the right thing. Let's look at another example. So let's take my same function j.

I want them to be improper fractions.

We're at an improper fraction.

So maybe M 47.

They're easy to think about oh, it's a little bit more than 8.

Another way to get to that is to remember the other rules for constructing environments.

If theta zero is 1.5 and theta one is 0, then the hypothesis function will look like this.

So we have two mixed numbers here, and like when you're adding mixed numbers, you can do it two ways.

And what's an improper fraction?

In the magenta situation, the thing in the absolute value sign was positive 2, because this was a situation where we assumed that the thing in the absolute value sign was a positive 2.

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