Setting the standard
Q I am a support teacher, with responsibility for a particular pupil.
Recently the class has been learning about standard form and my pupil has found this difficult. How can I help him?
A First, revisit powers of 10 so that he understands how to represent multiples of 10 in index form. Begin with representing 1,000 as 103 and 100 as 102. Ask him what he notices about each number and its index number, and demonstrate that 100 is 10 X 10 on the calculator. Do the same with 1,000.
Set up a table as I have shown, but with gaps, and work with your pupil to fill it correctly. Make sure he understands that the negative index means a reciprocal of 10 to the stated power. So 10-2 = 1Z102 = 1Z10x10 = 0.01.
Once again, look at the pattern in the zeros. Note that the powers of 10 are negative for small numbers (less than 1), positive for large numbers.
For students with writing difficulties, write each of the values in their different formats on separate cards and ask the student to sort them into the correct positions on the table - a simple but effective exercise which allows discussion as well.
On exam papers, candidates might be asked to write a number in the form a x 10n where 1 2 a . LESS THAN LESS THAN 10 and n is an integer. This is the general representation of standard form - make sure your pupil is familiar with it.
What it actually means is that a decimal number is changed into standard form while keeping its original value. o the non-posh expl
1 - that is multiplying by a power of 10 - would keep the original value the same.
Have a set of cards with various numbers written in standard form and some that aren't, and ask your student to decide which are in standard form (readers can email me if they would like a set). Having established that the student understands the form required, I suggest you demonstrate how to change between the two formats.
For example: change 3458 to standard form. For standard form we require "digit-dot-digits". The decimal point is to the right of the 8 at the moment. What is required is one digit, and then the decimal point. The new position of the point would be between 3 and 4. Draw a dotted line to show where the new position is to be and another dotted line in the old position. Underline each of the digits between the lines (here there are three). This tells us the power of 10 that is required in order to keep the value of the original number the same. So 3458 becomes 3.458 x 103.
Demonstrate this again with another example, where the decimal point is more obvious - eg 85.06. Using the dotted lines again gives us 85.06 = 8.506 x 101.
Next, tackle negative indices - eg 0.056. Once again, remind your student of the "digit-dot-digit" rule. The decimal point should be between the 5 and 6 (O is not in the range). Then, using the dotted line rule as before, ask your student if 0.056 is a large or small number. As it is small, we require a negative index. So 0.056 = 5.6 x 10-2 in standard form.
For a number such as 7.678, the decimal point is already in the correct position; the standard form is, therefore, 7.678 x 100, as the decimal point hasn't moved.
This method also works in reverse - that is, in changing from standard form to a decimal number. For example, take 5.58 x 104. This time, write 5.58 and draw the dotted line in the decimal point position. The 104 indicates that the decimal number is going to be bigger than 5.58 by four places, so draw four short lines, two beneath the two digits after the point (5 and 8), as seen in the diagram; then draw a dotted line vertically at the end to indicate the new position of the decimal point. Filling the blank spaces with zeros, the standard form for 5.58 x 104 becomes 55800.
For negative powers of 10, eg 3.8 x 10-3, write 3.8 as before, with the dotted line through the decimal point. The - 3 suggests that this is a small number, so this time underline spaces counting to the left of the dotted line. Draw another dotted line at the end of these and fill the spaces with zeros. Write another zero before the decimal point, as a standard notation.
So 3.8 x 10-3 becomes 0.0038.
Wendy Fortescue-Hubbard is a teacher and game inventor. She has been awarded a three-year fellowship by the National Endowment for Science, Technology and the Arts (NESTA) to spread maths to the masses.
Email your questions to Mathagony Aunt at firstname.lastname@example.org Or write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX