If 0 < x < 1, is it possible to write x as a terminating dec

Can someone explain me what is the meaning of terminating decimal.

A decimal number that has digits that do not go on forever.

Examples:

0.25 (it has two decimal digits)
0.123456789 (it has nine decimal digits)

In contrast a Recurring Decimal has digits that go on forever

Example of a Recurring Decimal: 1/3 = 0.333... (the 3 repeats forever)

I have a bit of difficulty in understanding the intended meaning of "is it possible" part of the question.

The answer can be yes and ofcourse no, but just that there is a possibility that the answer could be yes confuses me a bit. Had the question been framed like this " is x a terminating decimal?", then it would have been clearer. The use of the term "possible" makes it just a bit ambiguous.

Put up for guidance please.

The question basically asks: if x is written as a decimal will it be a terminating decimal?

Hope it's clear.

since the question asks "is it possible", wouldn't the answer be D since .5 is a terminating decimal and 24*.5=12, and 28*.5=24?

You misinterpret the question. The question asks: if x is written as a decimal will it be a terminating decimal? Thus the correct answer is C, not D.

we know that x is a proper positive fraction. we need to check whether x has powers of 5 or 2 in the denominator or not.

1. 24(x)=INT ---> \(x=Int/24\) if our integer is 3 then x is can be written as a terminating decimal otherwise x will be a non-terminating decimal

2. 28(x)=INT ----> same story here if our int is 7 then x can be written as a terminating decimal, otherwise x will be a non-terminating decimal

1+2 \(Int/3(2^3)=Int/7(2^2)\) -----> 7(4)Int=8(3)Int the expression has to be equal on both sides thus on the right hand side we need a 7 and on the right hand side we need a 3 and a two. We now know that our integer a terminating decimal because we can get rid of both 7 and 3 in the denominator.

C.

Hope it helps.

Bunuel
I am sorry
I can not get the combining statements.
We do not need to prove that n is a multiple of 7 ?
another question regarding the denominator, is it enough to be 2s or 5s to terminate X as decimal or 2s * 5s is a must.

From (1) we have that \(x=\frac{m}{24}=\frac{m}{2^3*3}\). If m is a multiple of 3, then 3 in the denominator will be reduced and x will be a terminating decimal.

Similarly, from (2) we have that \(x=\frac{n}{28}=\frac{n}{2^2*7}\). If n is a multiple of 7, then 7 in the denominator will be reduced and x will be a terminating decimal.

The answer to your other question is yes, if a fraction has only 2's or 5's in the denominator it'll terminate.

Check Terminating and Recurring Decimals Problems in our Special Questions Directory.

Hope it helps.

hi man

since 0<x<1, say x is a proper fraction ....

taking 2 statements together,

x cannot be 1/3, as 3 can divide 24, but cannot divide 28...
in the same line of reasoning, x cannot be 1/7, as 7 can divide 28, but cannot divide 24...
So, for both of the two statements to hold true, x cannot be 3 and/or 7, thus we are left with only 2s in the denominator. Sufficient ....

please say to me whether the reasoning is okay....

thanks in advance, man ...

Yes, x, when reduced to its simplest form must have 2 or 2^2 in the denominator.

OFFICIAL EXPLANATION
All terminating decimals share one property: the denominator of the decimal (written as a fraction) must contain only 2’s and 5’s as prime factors. For example, 0.25 written as a fraction is 1/4. The denominator contains only 2’s.

(1): INSUFFICIENT. For 24x to be an integer, the prime factors of 24 must completely cancel out the factors in the denominator of x. 24 = 23 × 3. There is no way to know which prime factors the denominator of x actually contains. If it contains a 3, the decimal will not terminate. If the denominator only contains 2’s, then it will terminate.

For example, if x is 1/2, 24x = 12, and the decimal terminates (0.5). If, however, x = 1/3, 24x = 8, but the decimal does not terminate (0.3333….). The statement is insufficient.

(2): INSUFFICIENT. For 28x to be an integer, the prime factors of 28 must completely cancel out the factors in the denominator of x. 28 = 22 × 7. There is no way to know which prime factors the denominator of x actually contains. If it contains a 7, the decimal will not terminate. If the denominator only contains 2’s, then it will terminate.

For example, if x is 1/2, 28x = 14, and the decimal terminates (0.5). If, however, x = 1/7, 28x = 4, but the decimal does not terminate (0.14285….). The statement is insufficient.

(1) & (2): SUFFICIENT. For both 24x to be an integer and 28x to be an integer, the denominator of x can only contain prime factors that both 24 and 28 share. The only prime factors that 24 and 28 share are 2 and 2(22). Therefore, the denominator of x must only contain 2’s, and the decimal must terminate.

The correct answer is C.

Hi, I have understood how you have explained the answer. But I solved the question in a different way. Please help me understand if my logic is right.

i. 24x = Int so x = 1/3 or x = 1/4. So NS

ii. 28x = Int so x = 1/4 or x = 1/7 NS

To combine, I did not list out the values of x. Instead, I did 28x - 24x = Int. So 4x = Int
x= 1/2 or x=1/4 in any case x will be non-terminating. Because there is no presence of 3 or 7 in the denominator.

So C

Yes, you are correct. For equation (1) though, x also can be 1/2, 1/6, 1/8, and 1/12. For equation (2), x can also be 1/2, 1/8, and 1/14. Additionally, for (1)+ (2), we obtain 4x = integer, which implies x = integer/4, which means that x will terminate because if the denominator of a fraction contains only 2's and/or 5's, then the fraction will be a terminating decimal.

Hi Bunuel , sorry but how does \(\frac{m}{n}=\frac{2*3}{7}\) prove that m is a multiple of 3? shouldnt it be M is a multiple of 6? i say this because if m is just a multiple of 3, then if we cross multiply, it will be 7(3)/6 = 3.5 which is not an integer.

Yes, \(\frac{m}{n}=\frac{2*3}{7}\) implies m is a multiple of 6. However, if it's a multiple of 6, then it must also be a multiple of 3. As we got in (1) if m is a multiple of 3, then x will be a terminating decimal because 3 in the denominator of \(x=\frac{m}{2^3*3}\) will get reduced.

Hope it's clear.