Calculations using scientific notation

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Probability

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Suppose I have a number 0.082 and I multiply it by 100. The number 0.082 moves two places to the left and reads; 8.2

Suppose then that same number is presented like this; [MATH](8.2\times10^-2)[/MATH]
If then this number is multiplied by [MATH](10^-2)[/MATH] it should read 0.082

Because the power is a negative the number goes smaller when multiplied.

Now look at this example;

Calculate with scientific notation;

[MATH](8.2\times(10^-2)-(5.4\times(10^-3)[/MATH]
If I now multiply [MATH](8.2)\times(10^-2)=(82)\times(10)\times(10^-3)=(82\times(10^-3)[/MATH]
I'm a little confused at this point because the decimal place has only moved 1 position, yet I'd of thought multiplying by [MATH](10^-2)[/MATH] would have moved 2 decimal places!

Before I move on would somebody please explain what I am not understanding.
 
Please use "^" for exponents.
The number 0.082 moves two places to the left and reads; 8.2 - right, not left
If then this number is multiplied by (10^−2) it should read 0.082 - how can it be 0.082 if it's the original number?
If I now multiply (8.2)×(10^−2)=(82)×(10)×(10^−3)=(82×(10^−3) - here the decimal point moved 1 place to the right because you replaced 10^−2 with 10^−3. You multiplied 8.2 by 10 and divided 10^−2 by 10, so the overall products stayed the same. If you multiply 8.2 by 10^−2, the result would be 0.082, as expected.
 
You could look at it this way

You could factor out [MATH]10^{-3}[/MATH]
[MATH]10^{-3}((8.2*10^{1})-5.4)[/MATH]
[MATH]10^{-3}(82-5.4)[/MATH]
Now distribute the [MATH]10^{-3}[/MATH]
[MATH](82*10^{-3})-(5.4*10^{-3})[/MATH]
 
Thank you. I do use "^" in laytex format but the software does not show it only the superscript being positive or negative, unless there is another way of writing it in the software.
If we refer to a number line, 1,2,3,4.... as shown here, -3, -2, -1. 0, 1, 2, 3, 4... and then enter my number 0.082 it is clearly smaller than number 1. If I multiply 0.082 by 100 the number moves by two decimal places to the left, which makes the number bigger, i.e. 8.2. The number appears to have moved to the right, but actually it has not, and the reason is that the "decimal point" never moves in any calculation.

However, what I'm not clear on here is (8.2 x 10^-2) = 8.2 x 10 and not being multiplied by 10^-2 as the number started with. As I see it (1) decimal place has been ignored otherwise the calculation would have used 8.2 x 10^-2 x 10^-3.

Where did the place value 10^-1 go when only (10) was carried forwards for the calculation!
 
In your problem

[MATH](8.2*10^{-2})-(5.4*10^{-3})[/MATH]
8.2 is ONLY BEING multiplied by [MATH]10^{-2}[/MATH] not by [MATH]10^{-3}[/MATH]
 
Probability said:
If I multiply 0.082 by 100 the number moves by two decimal places to the left, which makes the number bigger, i.e. 8.2. The number appears to have moved to the right, but actually it has not, and the reason is that the "decimal point" never moves in any calculation.
Click to expand...
I am very confused. On the number line 8.2 is to the right of 0.082. Please explain how you see it moving to the left. And what is the reference for the "decimal point" never moves in any calculation.? It's clearly false, 1.34*10 = 13.4. It moved to the right.
 
lev888 said:
I am very confused. On the number line 8.2 is to the right of 0.082. Please explain how you see it moving to the left. And what is the reference for the "decimal point" never moves in any calculation.? It's clearly false, 1.34*10 = 13.4. It moved to the right.
Click to expand...
While this is confusing, I was told by a mathematician that numbers going bigger move to the left, and numbers moving to the right go smaller.

13.4 Keep the decimal point in position, times by 10 = 134 now put that in decimal form; 134.0

Set your calculator up to show decimal places and try the calculation again, you'll see what I mean.
 
Probability said:
While this is confusing, I was told by a mathematician that numbers going bigger move to the left, and numbers moving to the right go smaller.

13.4 Keep the decimal point in position, times by 10 = 134 now put that in decimal form; 134.0

Set your calculator up to show decimal places and try the calculation again, you'll see what I mean.
Click to expand...
Ok, you were told something by someone. 1. That person could've been mistaken. 2. You might have misunderstood.
My point is that regardless of the source, if you can't explain the idea to someone else, maybe the idea is wrong.
So, if it's not confusing to you, please explain why you think 8.2 is to the left of 0.082 on the number line, where numbers increase from left to right.

13.4 Keep the decimal point in position, times by 10 = 134 now put that in decimal form; 134.0 - isn't this showing the decimal point moving???
 
lev888 said:
Ok, you were told something by someone. 1. That person could've been mistaken. 2. You might have misunderstood.
My point is that regardless of the source, if you can't explain the idea to someone else, maybe the idea is wrong.
So, if it's not confusing to you, please explain why you think 8.2 is to the left of 0.082 on the number line, where numbers increase from left to right.

13.4 Keep the decimal point in position, times by 10 = 134 now put that in decimal form; 134.0 - isn't this showing the decimal point moving???
Click to expand...
The decimal point never moves. If you think back to school days we were introduced to decimals.

Hundreds Tens Units [MATH]\cdot[/MATH]Tenths Hundredths Thousandths
1 [MATH]\cdot[/MATH] 3 2 5

The dot between 1 and 3 is positioned between units and tenths. Muliply by 10 gives; 13.25

This forum does not show this very well but on Amazon.co.uk this book which I have on page 1 of 1.1 digits and place value clearly shows what happens to the numbers and the decimal point.

Sometimes a picture paints a thousand words;

20200508_225515.jpg

If you look at the picture the title says multiply by 10, and the text shows it moves 1 place to the left The decimal point does not move but the unit value 1 moves left to the tens column and the 3 moves from the tens column to the units column.

Can you see it now!!!
 
Probability said:
Thank you. I do use "^" in laytex format but the software does not show it only the superscript being positive or negative, unless there is another way of writing it in the software.
Click to expand...
In LaTeX, to make anything more than a single digit into an exponent, you have to put it in braces. So 10^-1 looks like [MATH]10^-1[/MATH], while 10^{-1} looks like [MATH]10^{-1}[/MATH].

To see how others are successfully doing something in LaTeX, you can right-click on it and select "Show Math As ... TeX Commands".
 
Probability said:
The decimal point never moves. If you think back to school days we were introduced to decimals.

Hundreds Tens Units [MATH]\cdot[/MATH]Tenths Hundredths Thousandths
1 [MATH]\cdot[/MATH] 3 2 5

The dot between 1 and 3 is positioned between units and tenths. Muliply by 10 gives; 13.25

This forum does not show this very well but on Amazon.co.uk this book which I have on page 1 of 1.1 digits and place value clearly shows what happens to the numbers and the decimal point.

Sometimes a picture paints a thousand words;

View attachment 18643

If you look at the picture the title says multiply by 10, and the text shows it moves 1 place to the left The decimal point does not move but the unit value 1 moves left to the tens column and the 3 moves from the tens column to the units column.

Can you see it now!!!
Click to expand...
In this specific diagram I see that the digits shift to the left, yes. But I thought you were talking about the number line (see post #4). There the number would move to the right after multiplication by 10. Do we agree?
 
Maybe I did not explain it too well, which I'll accept full responsibility for, however it has come across, I was trying to explain it as my last post with the diagrams.

Getting back to my original problem [MATH](8.2)\times(10^{-2})=(8.2)\times(10)\times(10^{-3})[/MATH] If this [MATH](10^{-2})[/MATH] only represents 1 decimal place, when I thought it should represent two decimal places. Where did the 1 go that did not get mentioned in the math?

If you see my confusion multiplication by 10 = 1 d.p and not 2.d.p as I'd of expected from [MATH](10^{-2})[/MATH].
 
Probability said:
Getting back to my original problem [MATH](8.2)\times(10^{-2})=(8.2)\times(10)\times(10^{-3})[/MATH] If this [MATH](10^{-2})[/MATH] only represents 1 decimal place, when I thought it should represent two decimal places. Where did the 1 go that did not get mentioned in the math?

If you see my confusion multiplication by 10 = 1 d.p and not 2.d.p as I'd of expected from [MATH](10^{-2})[/MATH].
Click to expand...
How did you conclude that 10-2 represents 1 decimal place? Your example does not show the result of multiplication by 10-2. The result is 0.082, as expected the decimal point moved 2 positions to the left.
 
lev888 said:
How did you conclude that 10-2 represents 1 decimal place? Your example does not show the result of multiplication by 10-2. The result is 0.082, as expected the decimal point moved 2 positions to the left.
Click to expand...
I didn't, I said it should represent 2 decimal places. Read on as I believe I solved the problem last night...
 
I've always maintained that I am no mathematician, but I do like learning new ideas. When doing math on some calculators there are two functions, ENG and scientific notation. How factual this is I'm not 100% sure, but it seems when doing math in ENG mode and scientific mode, the power function shows 1.d.p difference.

What I mean by this is an answer in ENG mode might show[MATH](10^{-4})[/MATH] where scientific mode will show[MATH](10^{-3})[/MATH] and why that occurs must be something to do with the standards applied in mathematics subjects.

Now in my current example[MATH](8.2)\times(10^{-2})-(5.4\times(10^{-3})=[/MATH]
This takes a bit of understanding. Now [MATH](8.2\times(10^{-2})[/MATH] is I believe "Standard Notation", whereas [MATH](0.082)[/MATH] is decimal notation. If I were to use the product rule on [MATH](0.082)[/MATH]I would gain the result[MATH](8.2\times(10^{-2})[/MATH]
Now the author does not say what the math is applied to but just gives a question to be answered, in scientific notation. At this point I suspect that the student is then expected to know what standard of math to apply!

In this example[MATH](8.2)\times(10^{-2})[/MATH]I now understand that this is the standard form of[MATH](0.082)[/MATH] and because in my example I need the powers to be the same, initially I can just use the[MATH](8.2)[/MATH] from the first part of the math problem, and multiply that number by [MATH](10)[/MATH] to show;

[MATH](8.2)\times(10)[/MATH] which moves the decimals 1 decimal place to the left to show[MATH](82)\times(10^{-1})[/MATH]which in the math example appears to be shown as [MATH](10)[/MATH]only

At this point I seem to believe that I have a result of [MATH](82)\times(10^{-1})[/MATH] but the [MATH](10^{-1})[/MATH] is omitted. and shown as [MATH](82)\times(10^{-3})[/MATH] this is I assume to bring each standard notation to the same level of standard notation, i.e.[MATH](10^{-3})[/MATH]
At this point I'm looking at[MATH](8.2)\times(10^{-2})[/MATH] - [MATH](5.4)\times(10^{-3})=(82)\times(10^{-3})-(5.4)\times(10^{-3})[/MATH]
This must be the way the author allows the powers to become the same value for each part of the math problem!

So up to this point I'm saying that;

[MATH](8.2)\times(10^{-2})-(5.4)\times(10^{-3})=[/MATH][MATH](82)\times(10^{-3})-(5.4)\times(10^{-3})[/MATH]but really I still see this as;

[MATH](8.2)\times(10^{-1})[/MATH]because it was only ever multiplied by (10)

Anyway now the powers are the same, so I can now say;

[MATH](82)-(5.4)\times(10^{-3})=76.6\times(10^{-3})=[/MATH]
[MATH]77\times(10^{-3})[/MATH](to 2 s.f) ((This must be the ENG notation))

Now the scientific notation;

We must not forget that we started by multiplying [MATH](10^{-2})[/MATH] by [MATH](10)[/MATH] only so I think pay back must occur.

which must be the ENG notation, and this then must be the scientific notation, hence

[MATH](77)\times(10^{-3})=(7.7)\times(10)\times(10^{-3})=(7.7\times(10^{-2})[/MATH] so that;

[MATH](8.2\times(10^{-2})-(5.4)\times(10^{-3})=(7.7)\times(10^{-2})[/MATH] (to 2 s.f.)

Feedback would be greatly appreciated. This has been a learning curve over the last 24 hrs.
 
Probability said:
I've always maintained that I am no mathematician, but I do like learning new ideas. When doing math on some calculators there are two functions, ENG and scientific notation. How factual this is I'm not 100% sure, but it seems when doing math in ENG mode and scientific mode, the power function shows 1.d.p difference.

What I mean by this is an answer in ENG mode might show[MATH](10^{-4})[/MATH] where scientific mode will show[MATH](10^{-3})[/MATH] and why that occurs must be something to do with the standards applied in mathematics subjects.

Now in my current example[MATH](8.2)\times(10^{-2})-(5.4\times(10^{-3})=[/MATH]
This takes a bit of understanding. Now [MATH](8.2\times(10^{-2})[/MATH] is I believe "Standard Notation", whereas [MATH](0.082)[/MATH] is decimal notation. If I were to use the product rule on [MATH](0.082)[/MATH]I would gain the result[MATH](8.2\times(10^{-2})[/MATH]
Now the author does not say what the math is applied to but just gives a question to be answered, in scientific notation. At this point I suspect that the student is then expected to know what standard of math to apply!

In this example[MATH](8.2)\times(10^{-2})[/MATH]I now understand that this is the standard form of[MATH](0.082)[/MATH] and because in my example I need the powers to be the same, initially I can just use the[MATH](8.2)[/MATH] from the first part of the math problem, and multiply that number by [MATH](10)[/MATH] to show;

[MATH](8.2)\times(10)[/MATH] which moves the decimals 1 decimal place to the left to show[MATH](82)\times(10^{-1})[/MATH]which in the math example appears to be shown as [MATH](10)[/MATH]only

At this point I seem to believe that I have a result of [MATH](82)\times(10^{-1})[/MATH] but the [MATH](10^{-1})[/MATH] is omitted. and shown as [MATH](82)\times(10^{-3})[/MATH] this is I assume to bring each standard notation to the same level of standard notation, i.e.[MATH](10^{-3})[/MATH]
At this point I'm looking at[MATH](8.2)\times(10^{-2})[/MATH] - [MATH](5.4)\times(10^{-3})=(82)\times(10^{-3})-(5.4)\times(10^{-3})[/MATH]
This must be the way the author allows the powers to become the same value for each part of the math problem!

So up to this point I'm saying that;

[MATH](8.2)\times(10^{-2})-(5.4)\times(10^{-3})=[/MATH][MATH](82)\times(10^{-3})-(5.4)\times(10^{-3})[/MATH]but really I still see this as;

[MATH](8.2)\times(10^{-1})[/MATH]because it was only ever multiplied by (10)

Anyway now the powers are the same, so I can now say;

[MATH](82)-(5.4)\times(10^{-3})=76.6\times(10^{-3})=[/MATH]
[MATH]77\times(10^{-3})[/MATH](to 2 s.f) ((This must be the ENG notation))

Now the scientific notation;

We must not forget that we started by multiplying [MATH](10^{-2})[/MATH] by [MATH](10)[/MATH] only so I think pay back must occur.

which must be the ENG notation, and this then must be the scientific notation, hence

[MATH](77)\times(10^{-3})=(7.7)\times(10)\times(10^{-3})=(7.7\times(10^{-2})[/MATH] so that;

[MATH](8.2\times(10^{-2})-(5.4)\times(10^{-3})=(7.7)\times(10^{-2})[/MATH] (to 2 s.f.)

Feedback would be greatly appreciated. This has been a learning curve over the last 24 hrs.
Click to expand...
In general,

ENG mode should display exponents that are divisible by 3 e.g.

... 10-9, 10-6, 10-3, 103, 106, 109, 10-12 ......
 
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