Let x = 0.99...
Then 10x = 9.99...
So 10x - x = 9.99... - 0.99... = 9 [the strings after the decimal point are infinite so they are the same]
ie 9x = 9
So x = 1
Another approach is to say that given any number k, however small, it is possible to find a number n such that 0.999... to n digits is nearer to 1 than k. This is known as the limiting value of 0.99...
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actually anything divided by itself equals 1 .99999... (repeated infinity) is as close as you can get to 1 perhaps your calculator is rounding on the last digit
The answer is simply 99999999.
It is equal to two, just as 0.9999 repeated is equal to one. Here is a proof: 1/3=0.33333 repeating (1/3)*3=0.3333 repeating * 3 1=0.9999 repeating Now, adding one to both sides also means that 1.9999 repeating equals 2.
1+3+5+5+17+19 they are 5 numbers 5 is repeated
Yes.