How do you write -5/8 as a decimal?

2 Answers
Sep 6, 2017

-0.6250.625

Explanation:

color(blue)("Introduction to a way of thinking about this problem")Introduction to a way of thinking about this problem
As the 5 is less than 8 it may gives you a problem.

What follows is a bit of a cheat (not really)

Using the principle that we can change the 5 or any other number any way we wish as long as we include a way to turn the change back to its original value.

If we so wished we could change it as follows:

As given: 5/8->5xx1/8585×18

=50xx1/10xx1/8=50×110×18

=500xx1/100xx1/8=500×1100×18

=5000xx1/1000xx1/8=5000×11000×18

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Answering the question")Answering the question

Ignoring the negative for now.

You accumulate the adjustment in this approach

color(green)("5 is less than 8 so change it to 50")5 is less than 8 so change it to 50

50xx1/8=color(magenta)(6)" remainder 2 "50×18=6 remainder 2

color(green)("2 is less than 8 so change it to 20")2 is less than 8 so change it to 20

20xx1/8=color(magenta)(2)" remainder 4 "20×18=2 remainder 4

color(green)("4 is less than 8 so change it to 40")4 is less than 8 so change it to 40

40xx1/8=color(magenta)(5)" remainder 0"40×18=5 remainder 0

Putting our numbers together we have color(magenta)(625)625
................................................................................................

color(white)()

However, we had color(green)("3 adjustments")3 adjustments so we multiply by three lots of 1/10110 to bring it back to the value it should be.

625xx1/10xx1/10xx1/10" = "0.625625×110×110×110 = 0.625

Putting the negative back in we have -0.6250.625

Feb 12, 2018

The same thing as the other solution but a neater presentation.

-0.625

Explanation:

For now forget about the negative. We can out it back at the end

Write 5 as 50xx1/1050×110

Start point ->50xxcolor(green)(1/10)50×110
color(magenta)(6)xx8 ->color(white)("dddd")ul(48larr" Subtract")
ul(color(white)(~~~~~~~~~.)2 color(white)(~~~~~~~~~))

color(white)("d")
color(white)("~~~~~~~~~~~~.")20xxcolor(green)(1/10)larr" remainder of 2 changed"
color(magenta)(2)xx8->color(white)("dddd")ul(16larr" Subtract")
ul(color(white)("dddddddddddd")4color(white)("ddddddddddd"))

color(white)("d")
color(white)("ddddddddddd")40xxcolor(green)(1/10) larr" Remainder of 4 changed"
color(magenta)(5)xx8->color(white)("dddd")ul(40larr" Subtract")
color(white)("dddddddddddd")0 larr" Stop"

So for just the numbers we have:

color(white)("d")
color(magenta)(625)color(green)(xx1/10xx1/10xx1/10)

color(magenta)(0.625)

Now we put the 'negative state' back giving

color(magenta)(-0.625)