Calculations with very big or small numbers can be made easier by converting numbers in and out of standard form.

Standard form, or standard index form, is a system of writing numbers which can be particularly useful for working with very large or very small numbers. It is based on using powers of 10 to express how big or small a number is.

We are looking at standard form. So we'll first focus on what's the point of standard form and what is it. We'll see how it can be applied to both big numbers and small numbers, how we can multiply two standard form numbers together and then finally do some exam star questions to finish off. So what is standard form and what's the point of standard form? Well, standard form is really useful if we want to write down really big or really small numbers.

Imagine if we're talking about say, the number of stars in Moka Bay. Well, it's roughly 100 billion, right? So imagine we've measured this number of stars and it's 113,000,000,000 stars. So it's one one three with nine zeros at the end. Will it be a real hassle to have to write down this long string of zeros every single time we wanted to reference this value?

Right, imagine we've got a scientific paper. We don't want half of our paper to just be zeros. It would be much nicer if we can condense this down and represent this value in a much shorter format. And what we can do is we can use standard form to achieve that. So there's one rule in standard form and that is that we have a number at the start and that number has to be between one and ten.

We then take this original number and we times it by some scale factor and we make it either much, much bigger or much smaller, depending on our needs, depending on what the original number is that we want to represent. So in this case, let's focus on this. We've got this massive number and we need to first think of a number between one and ten that can represent our digits. Well, in this case it's going to be 1.3, right? We'll see why in a second.

Where if we tried any other number to represent these digits, if we tried, say, 11.3, well, 11.3 is bigger than ten, right? It's not between one and ten, so that wouldn't be correct. If we tried 0.13 again, that wouldn't be correct. We know that 0.3 is not between one and ten. The correct answer is going to be 1.3 for that first value.

So this is a kind of base value and this is our scale factor. So what we now do, now that we've got our kind of base value is we work out what is the scale factor, how much do we need to times 1.13 by to convert it into 113,000,000,000? Well, if we look at where the decimal point is at the moment in this original number, it's at the end here, right? There's actually a decimal point there. We just don't bother putting it in.

And at the moment we've got our decimal point in between these first two ones, our decimal point is here at the moment in a number. So we need to convert from this tiny number with the decimal point here to this massive number with the decimal point here. And so we just work out how many columns we need to shift our decimal point across by to make this number as big as it needs to be. Right? So if we look at the number of jumps we need to make it's going to be 123-4567, 8910, eleven spaces we need to jump and so we're basically timing it by 1011 times over.

We're timing something by ten and times it by ten again and ten again. Eleven times it's just times it by ten power of eleven actually put that in there. This is how we can represent our number in standard form. That would be part of our solution. So that's kind of half of big numbers covered actually just in the definition of standard form.

We can also work back the other way so we can take a number in standard form and we can represent it as an ordinary number. Let's say we take 52 times ten to the times ten to the nine and we want to represent this as a standard form number. Let's say this is actually 5.0 that'd be our correct standard format and we want to convert this into an ordinary number. So there's a number with no standard form in it anymore. Well we know we have to start off with our decimal point between the five and the two and we need to shift it down nine spaces, right, nine columns thereby making the number ten to nine times bigger.

And so if we do that, if we shift it down nine spaces that's 123-45-6789 it's now going to go there and all we do is we fill in the gaps with zero. So every single time we have an empty column we just put a zero there. And this is our new number, right? Our new number is 5200.

We put our commas if you wanted t that's going to be 5,200,000,000 and that's going to be our answer to you.

We can also do small numbers. So again imagine we're talking about say the width of a proton that's around eight times ten to -14 metres you often see this in textbooks and things talking about protons. Well again this is much nicer way of doing this. If we wanted to convert this to an ordinary number actually if you want to convert this to an ordinary number all we do is very similar to what we did here right we start off with this is basically 8.0 times ten to -14 we start off with a decimal point. Here we know that we need to move a decimal point to the left in order to make our number ten to 14 times smaller.

When we look at how many times we need to make that jump in this case it's 14 times we have to jump across 14 columns and so if I give myself some room 8.0 if we jump across 14 columns, that's 123-45, 67, 89, 10, 11, 12, 13, 14. Should have picked a better number but there we go. And we can fill in our gaps with zeros like we always do, because we can't any other number than zero in here, that's going to be a new number, it's going to be all the way down to eight there. And likewise we can go the other way as well, right? We can take a number that's an ordinary number and convert it back into standard form.

Let's find a look at multiplication. So if we've got say, two numbers both in standard form, we could have say, I don't know, let's keep it nice, three times ten to five times four times ten to the seven. Well, we've just got three sorry, four different terms all times in each other together, right? If we've got a bunch of terms all times in each other together, it doesn't matter what way around we do the modification in. Let's say we've got say, two times three times five.

Well that's the exact same as doing five times two times three, right? And it's the exact same as doing three times five times two or whatever we want to do. In all cases we're going to get 30, right? Every single time, no matter what way around we do it, we always end up getting 30. And so we can use this basic principle to make this bit nicer for ourselves, right?

We can move our numbers together and move our powers together and say we can write this as three times four, times ten to five times ten to the seven windows. Three times four is going to give us twelve and ten to the five times ten to the seven is going to give us ten to the twelve. And it would be very tempting to leave it there. Normally when we answer these kind of questions we need to put our answer in standard form as well. And then we can see this twelve is not between one and ten.

And so we look at this, we say, well, how do we take these digits of one and two and make a number that's between one and ten? Well, that's going to be 1.2, that's going to be our initial value and then our scale factor. If we've made this number ten times smaller we've got to make our scale factor ten times bigger so that the overall value of our final answer doesn't actually change. And so instead of ten to twelve, it's going to be ten to 13 and that'll be our final answer. Let's do a couple of exam style questions to finish off with.

So we want to write 3.8 times ten to the minus five as an ordinary number. So I'll just write down here 3.8 times ten to the minus five. In this case we're going to be shifting our decimal point up to the left to make our number move down to the right. And so I just take this decimal point I might write out down here, 3.8, and I shift this five spaces to the left, three, four, five. So that's going to be my new position.

I fill in the gaps with zeros. And so our final answer is going to be 0.5 million in storm. Well, 5 million a million. This is ten times ten times ten times ten times ten times ten, right? It's one with 60 at the end of it.

And we can represent that in standard form as just ten to the power of six. And again, rather than just 1 million, it's 5 million. So it's five lots of 1 million. So this is 1 million here.

Finally, calculate 6.3 times five, times 2.5, times ten to the minus T out down here, times ten to the five times T, times ten to the minus T. And again, we use the same principle as the previous page, right, we've got four different times here, all times in each other. Together we can rearrange this as we like. So I'm going to put my numbers at the front, 6.3 times 2.5 and my powers at the back, times ten to the five, times ten to the minus two. I can work this out first, assuming this is a non calculated paper, I can do 6.3 times 2.5.

And if I do that, I have to remember what column I'm starting in. I can do both ways, actually. So I do zero three times zero five. That's going to give me zero here first in my mind. And I start in the column over.

That's going to give me 1515 and then I can do zero three times T, that's going to give me six plus the one is seven, and then do six times zero five. And I'm going to start on this column now, that's going to give me 30. So I carry the 30 there and then do six times two. That's twelve plus the point 75. That's probably easier than we needed to see.

We can take us on this one as well. Ten to the five, times ten to the minus t, that's going to give us ten to the power of three. All we're doing again is we're just adding those two powers together and so attempting to write down our answer as 15.75 times ten to the three. What we see is we need our answer in standard form at the moment, this is not in standard form, this is not between one and ten. I have to make this first number ten times smaller and I don't have to make this second number ten times bigger to compare to standard form.

So we end up with zero 1.575 times ten, the power of four. And that is our final answer. And at the end of step four.

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