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

2 Answers
Sep 6, 2017

#-0.625#

Explanation:

#color(blue)("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->5xx1/8#

#=50xx1/10xx1/8#

#=500xx1/100xx1/8#

#=5000xx1/1000xx1/8#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("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")#

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

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

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

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

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

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

#color(white)()#

However, we had #color(green)("3 adjustments")# so we multiply by three lots of #1/10# to bring it back to the value it should be.

#625xx1/10xx1/10xx1/10" = "0.625#

Putting the negative back in we have #-0.625#

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 #50xx1/10#

Start point #->50xxcolor(green)(1/10)#
#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)#