My Math Forum Confused about dividing and multiplying scientific notations.

 Algebra Pre-Algebra and Basic Algebra Math Forum

 May 17th, 2015, 09:29 AM #1 Member   Joined: Mar 2015 From: earth Posts: 52 Thanks: 0 Confused about dividing and multiplying scientific notations. (3.3×10​^−2​​)×(4.0×10​^−2​​)= ? 1.32^-3 why it's negative three? please explain Last edited by decimate; May 17th, 2015 at 09:38 AM.
 May 17th, 2015, 10:12 AM #2 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 $\displaystyle \color{blue}3.3\cdot10^{-2}\cdot4\cdot10^{-2}=(3.3\cdot4)\cdot(10^{-2}\cdot10^{-2})=13.2\cdot10^{-4}=13.2\cdot10^{-1}\cdot10^{-3}=1.32\cdot10^{-3}$
May 17th, 2015, 10:27 AM   #3
Member

Joined: Mar 2015
From: earth

Posts: 52
Thanks: 0

Quote:
 Originally Posted by aurel5 $\displaystyle \color{blue}3.3\cdot10^{-2}\cdot4\cdot10^{-2}=(3.3\cdot4)\cdot(10^{-2}\cdot10^{-2})=13.2\cdot10^{-4}=13.2\cdot10^{-1}\cdot10^{-3}=1.32\cdot10^{-3}$
I can't comprehend the thing greater than 10 something?

 May 17th, 2015, 11:30 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 RULE: x^(-p) = 1 / x^p : TATTOO that on your wrist! 10^(-2) = 1/10^2 = 1/100 3.3(1/100) * 4(1/100) = 13.2 / 10000 = .00132
May 18th, 2015, 04:04 AM   #5
Member

Joined: Mar 2015
From: earth

Posts: 52
Thanks: 0

Quote:
 Originally Posted by Denis RULE: x^(-p) = 1 / x^p : TATTOO that on your wrist! 10^(-2) = 1/10^2 = 1/100 3.3(1/100) * 4(1/100) = 13.2 / 10000 = .00132
I can't still understand I think I have dyscalculia.

May 18th, 2015, 04:16 AM   #6
Senior Member

Joined: Jul 2014
From: भारत

Posts: 1,178
Thanks: 230

Quote:
 Originally Posted by decimate I can't still understand I think I have dyscalculia.
It is just an exponential rule.
If a number has a negative power, it is equivalent to its multiplicative inverse (reciprocal) with the same power in positive.

$\displaystyle x^{-m}=\frac{1}{x^{m}} \\ \text{Example: }3^{-3}=\frac{1}{3^{3}}=\frac{1}{9}$

 May 18th, 2015, 05:51 AM #7 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timey-wimey stuff. For those who are unaware of what discalculia is, see here. I have met a person with this problem, but it is extremely rare. Most people that claim to have it simply have a hard time with Math. Have you been diagnosed? -Dan
May 19th, 2015, 01:28 AM   #8
Senior Member

Joined: Apr 2014
From: Glasgow

Posts: 2,157
Thanks: 732

Math Focus: Physics, mathematical modelling, numerical and computational solutions
Decimate, let's take a look at each part of Aurel's working step-by-step

Quote:
 Originally Posted by aurel5 $\displaystyle \color{blue}3.3\cdot10^{-2}\cdot4\cdot10^{-2}=(3.3\cdot4)\cdot(10^{-2}\cdot10^{-2})=13.2\cdot10^{-4}$
Firstly, some people use a $\displaystyle \cdot$ symbol to mean "multiply". It's actually really common practise when you do higher level mathematics. Just to avoid confusion, I'm going to use the multiplication symbol $\displaystyle \times$ instead.

At this part, Aurel multiplied the two "first bits " together (called significands) to get 13.2 and then multiplied the "second parts together" (called powers of ten. The number on each 10 is called an exponent or a power.) to get $\displaystyle 10^{-4}$

$\displaystyle 13.2 \times 10^{-4}$ is basically the answer. However, it is not in standard form. For a number to be in standard form, its significand must be between 1 and 10, but not exactly equal to 10 (i.e. 9.999999999 is allowed, but not 10). At the moment the significant is 13.2, which is greater than 10. This is where we move the to the next part of Aurel's calculation.

Quote:
 Originally Posted by aurel5 $\displaystyle 13.2\cdot10^{-4}=13.2\cdot10^{-1}\cdot10^{-3}=1.32\cdot10^{-3}$
Aurel has converted the number to standard form. He did this by factorising out a factor of 0.1 (the same as $\displaystyle 10^{-1}$) from the $\displaystyle 10^{-4}$ and then multiply this by 13.2 to get 1.32, which now follows the rule for standard form.

When doing these, I like to think of it as like a "sliding scale"... if you reduce one thing by a factor of 10, then you must increase the other thing by a factor of 10 to balance it out and vice versa. For example, consider all of the numbers below. They are all the same:

$\displaystyle 132 \times 10^{-5}$
$\displaystyle 13.2 \times 10^{-4}$
$\displaystyle 1.32 \times 10^{-3}$
$\displaystyle 0.132 \times 10^{-2}$
$\displaystyle 0.0132 \times 10^{-1}$
$\displaystyle 0.00132 \times 10^{0} = 0.00132$

So by moving the decimal place by 1 digit to the left as we move down the list (which makes the significant smaller), the exponent increases by 1 (which makes the second part bigger).

May 19th, 2015, 01:33 AM   #9
Member

Joined: Mar 2015
From: earth

Posts: 52
Thanks: 0

Quote:
 Originally Posted by topsquark For those who are unaware of what discalculia is, see here. I have met a person with this problem, but it is extremely rare. Most people that claim to have it simply have a hard time with Math. Have you been diagnosed? -Dan

Not yet

 May 19th, 2015, 05:11 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 See if you can follow this: Code: x-axis <........-3...........-2..........-1.........0.........1..........2..........3.....> 10^n 10^-3 10^-2 10^-1 10^0 10^1 10^2 10^3 result 1/1000 1/100 1/10 1 10 100 1000 same as 1/10^3 1/10^2 1/10^1 Take the "-3" column: can you kinda see why 10^-3 = 1/10^3 ?

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mathbalarka New Users 10 May 15th, 2013 08:22 AM sallyyy Algebra 4 June 30th, 2011 05:34 PM badrush Algebra 3 January 31st, 2010 03:51 PM gear2d Applied Math 5 September 21st, 2008 07:06 AM johnny Algebra 1 August 18th, 2007 02:37 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top