Changing number bases for expansion of fractions
Okay, so my question is, say if I have $141_{10}$ and I convert it into hexadecimal, I get $8D_{16}$ but if it I have $0.141_{10}$ and I convert it to hexadecimal, I get $0.24189374BC6A7EF9E$, why is there this discrepancy?
number-theory
edited Dec 2 '18 at 13:48
saulspatz
14k21329
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
add a comment |
Okay, so my question is, say if I have $141_{10}$ and I convert it into hexadecimal, I get $8D_{16}$ but if it I have $0.141_{10}$ and I convert it to hexadecimal, I get $0.24189374BC6A7EF9E$, why is there this discrepancy?
number-theory
edited Dec 2 '18 at 13:48
saulspatz
14k21329
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
1
Are you asking why you don't get $.8d$ in hex? If so, it's because $.8d = {8dover 256_{10}}$ not ${8dover 1000_{10}}$
– saulspatz
Dec 2 '18 at 13:47
Okay, but then how am I supposed to change such values?
– AryanSonwatikar
Dec 2 '18 at 13:51
add a comment |
Okay, so my question is, say if I have $141_{10}$ and I convert it into hexadecimal, I get $8D_{16}$ but if it I have $0.141_{10}$ and I convert it to hexadecimal, I get $0.24189374BC6A7EF9E$, why is there this discrepancy?
number-theory
edited Dec 2 '18 at 13:48
saulspatz
14k21329
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
Okay, so my question is, say if I have $141_{10}$ and I convert it into hexadecimal, I get $8D_{16}$ but if it I have $0.141_{10}$ and I convert it to hexadecimal, I get $0.24189374BC6A7EF9E$, why is there this discrepancy?
number-theory
number-theory
edited Dec 2 '18 at 13:48
saulspatz
14k21329
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
edited Dec 2 '18 at 13:48
saulspatz
14k21329
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
edited Dec 2 '18 at 13:48
saulspatz
14k21329
edited Dec 2 '18 at 13:48
saulspatz
14k21329
edited Dec 2 '18 at 13:48
saulspatz
14k21329
14k21329
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
asked Dec 2 '18 at 13:28
AryanSonwatikar
31312
31312
1
Are you asking why you don't get $.8d$ in hex? If so, it's because $.8d = {8dover 256_{10}}$ not ${8dover 1000_{10}}$
– saulspatz
Dec 2 '18 at 13:47
Okay, but then how am I supposed to change such values?
– AryanSonwatikar
Dec 2 '18 at 13:51
add a comment |
1
Are you asking why you don't get $.8d$ in hex? If so, it's because $.8d = {8dover 256_{10}}$ not ${8dover 1000_{10}}$
– saulspatz
Dec 2 '18 at 13:47
Okay, but then how am I supposed to change such values?
– AryanSonwatikar
Dec 2 '18 at 13:51
1
1
Are you asking why you don't get $.8d$ in hex? If so, it's because $.8d = {8dover 256_{10}}$ not ${8dover 1000_{10}}$
– saulspatz
Dec 2 '18 at 13:47
Are you asking why you don't get $.8d$ in hex? If so, it's because $.8d = {8dover 256_{10}}$ not ${8dover 1000_{10}}$
– saulspatz
Dec 2 '18 at 13:47
Okay, but then how am I supposed to change such values?
– AryanSonwatikar
Dec 2 '18 at 13:51
Okay, but then how am I supposed to change such values?
– AryanSonwatikar
Dec 2 '18 at 13:51
add a comment |
1 Answer
1
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oldest
votes
Suppose that the hexadecimal fraction is $f=.a_1a_2a_3..._{16}$ Then $$16f = a_1.a_2a_3.._{16}.$$ So, if we multiply the fraction by $16,$ the integer part of the result is the first hexadecimal digit. $$16cdot.141=2.256,$$ so the first hex digit after the "hexadecimal point" is $2.$ To get the second digit, ignore the integer part of the last result, and multiply by $16$ again.$$16cdot.256=4.096,$$ so the second digit is $4$. Then $$16cdot.096=1.536$$ and so on. $$boxed{.141_{10}=.241..._{16}}$$
answered Dec 2 '18 at 14:58
saulspatz
14k21329
add a comment |
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Suppose that the hexadecimal fraction is $f=.a_1a_2a_3..._{16}$ Then $$16f = a_1.a_2a_3.._{16}.$$ So, if we multiply the fraction by $16,$ the integer part of the result is the first hexadecimal digit. $$16cdot.141=2.256,$$ so the first hex digit after the "hexadecimal point" is $2.$ To get the second digit, ignore the integer part of the last result, and multiply by $16$ again.$$16cdot.256=4.096,$$ so the second digit is $4$. Then $$16cdot.096=1.536$$ and so on. $$boxed{.141_{10}=.241..._{16}}$$
answered Dec 2 '18 at 14:58
saulspatz
14k21329
add a comment |
Suppose that the hexadecimal fraction is $f=.a_1a_2a_3..._{16}$ Then $$16f = a_1.a_2a_3.._{16}.$$ So, if we multiply the fraction by $16,$ the integer part of the result is the first hexadecimal digit. $$16cdot.141=2.256,$$ so the first hex digit after the "hexadecimal point" is $2.$ To get the second digit, ignore the integer part of the last result, and multiply by $16$ again.$$16cdot.256=4.096,$$ so the second digit is $4$. Then $$16cdot.096=1.536$$ and so on. $$boxed{.141_{10}=.241..._{16}}$$
answered Dec 2 '18 at 14:58
saulspatz
14k21329
add a comment |
Suppose that the hexadecimal fraction is $f=.a_1a_2a_3..._{16}$ Then $$16f = a_1.a_2a_3.._{16}.$$ So, if we multiply the fraction by $16,$ the integer part of the result is the first hexadecimal digit. $$16cdot.141=2.256,$$ so the first hex digit after the "hexadecimal point" is $2.$ To get the second digit, ignore the integer part of the last result, and multiply by $16$ again.$$16cdot.256=4.096,$$ so the second digit is $4$. Then $$16cdot.096=1.536$$ and so on. $$boxed{.141_{10}=.241..._{16}}$$
answered Dec 2 '18 at 14:58
saulspatz
14k21329
Suppose that the hexadecimal fraction is $f=.a_1a_2a_3..._{16}$ Then $$16f = a_1.a_2a_3.._{16}.$$ So, if we multiply the fraction by $16,$ the integer part of the result is the first hexadecimal digit. $$16cdot.141=2.256,$$ so the first hex digit after the "hexadecimal point" is $2.$ To get the second digit, ignore the integer part of the last result, and multiply by $16$ again.$$16cdot.256=4.096,$$ so the second digit is $4$. Then $$16cdot.096=1.536$$ and so on. $$boxed{.141_{10}=.241..._{16}}$$
answered Dec 2 '18 at 14:58
saulspatz
14k21329
answered Dec 2 '18 at 14:58
saulspatz
14k21329
answered Dec 2 '18 at 14:58
saulspatz
14k21329
answered Dec 2 '18 at 14:58
saulspatz
14k21329
14k21329
add a comment |
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1
Are you asking why you don't get $.8d$ in hex? If so, it's because $.8d = {8dover 256_{10}}$ not ${8dover 1000_{10}}$
– saulspatz
Dec 2 '18 at 13:47
Okay, but then how am I supposed to change such values?
– AryanSonwatikar
Dec 2 '18 at 13:51