If the fraction answer is 2 and the decimal answer is 1.9 repeating and it says to compare and make sure theyr the same is the repeating decimal rounded to be the same as the 2?
I think you're being asked to show that,\[1.9999999...=2\]This is one of those things where the introduction of an infinity of something gives you something (here, the infinitely repeating 9's) that is true, but counter-intuitive. To see this, let \[x=1.99999999...\]then\[10x=19.9999999...\]If you subtract x from 10x, you get\[10x-x=9x=19.9999999...-1.99999999...=19-1=18\]In the middle-ish step, we've used the fact that you're taking away the infinity of repeating 9's from each number (this is proved a lot better if using sigma notation and the definition of numbers in base 10, but I think this gets the message across). So now you have \[9x=18 \rightarrow x=\frac{18}{9}=2\]But we said that \[x=1.9999999...\]So\[1.9999999...=2\]If you're required to show a 'proper' proof, let me know. Hope I've explained it well :)
no i dont need proper proof thanx for the help!(;
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