-0.016 738 891 601 562 494 59 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 494 59(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.016 738 891 601 562 494 59(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.016 738 891 601 562 494 59| = 0.016 738 891 601 562 494 59


2. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

3. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 494 59.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.016 738 891 601 562 494 59 × 2 = 0 + 0.033 477 783 203 124 989 18;
  • 2) 0.033 477 783 203 124 989 18 × 2 = 0 + 0.066 955 566 406 249 978 36;
  • 3) 0.066 955 566 406 249 978 36 × 2 = 0 + 0.133 911 132 812 499 956 72;
  • 4) 0.133 911 132 812 499 956 72 × 2 = 0 + 0.267 822 265 624 999 913 44;
  • 5) 0.267 822 265 624 999 913 44 × 2 = 0 + 0.535 644 531 249 999 826 88;
  • 6) 0.535 644 531 249 999 826 88 × 2 = 1 + 0.071 289 062 499 999 653 76;
  • 7) 0.071 289 062 499 999 653 76 × 2 = 0 + 0.142 578 124 999 999 307 52;
  • 8) 0.142 578 124 999 999 307 52 × 2 = 0 + 0.285 156 249 999 998 615 04;
  • 9) 0.285 156 249 999 998 615 04 × 2 = 0 + 0.570 312 499 999 997 230 08;
  • 10) 0.570 312 499 999 997 230 08 × 2 = 1 + 0.140 624 999 999 994 460 16;
  • 11) 0.140 624 999 999 994 460 16 × 2 = 0 + 0.281 249 999 999 988 920 32;
  • 12) 0.281 249 999 999 988 920 32 × 2 = 0 + 0.562 499 999 999 977 840 64;
  • 13) 0.562 499 999 999 977 840 64 × 2 = 1 + 0.124 999 999 999 955 681 28;
  • 14) 0.124 999 999 999 955 681 28 × 2 = 0 + 0.249 999 999 999 911 362 56;
  • 15) 0.249 999 999 999 911 362 56 × 2 = 0 + 0.499 999 999 999 822 725 12;
  • 16) 0.499 999 999 999 822 725 12 × 2 = 0 + 0.999 999 999 999 645 450 24;
  • 17) 0.999 999 999 999 645 450 24 × 2 = 1 + 0.999 999 999 999 290 900 48;
  • 18) 0.999 999 999 999 290 900 48 × 2 = 1 + 0.999 999 999 998 581 800 96;
  • 19) 0.999 999 999 998 581 800 96 × 2 = 1 + 0.999 999 999 997 163 601 92;
  • 20) 0.999 999 999 997 163 601 92 × 2 = 1 + 0.999 999 999 994 327 203 84;
  • 21) 0.999 999 999 994 327 203 84 × 2 = 1 + 0.999 999 999 988 654 407 68;
  • 22) 0.999 999 999 988 654 407 68 × 2 = 1 + 0.999 999 999 977 308 815 36;
  • 23) 0.999 999 999 977 308 815 36 × 2 = 1 + 0.999 999 999 954 617 630 72;
  • 24) 0.999 999 999 954 617 630 72 × 2 = 1 + 0.999 999 999 909 235 261 44;
  • 25) 0.999 999 999 909 235 261 44 × 2 = 1 + 0.999 999 999 818 470 522 88;
  • 26) 0.999 999 999 818 470 522 88 × 2 = 1 + 0.999 999 999 636 941 045 76;
  • 27) 0.999 999 999 636 941 045 76 × 2 = 1 + 0.999 999 999 273 882 091 52;
  • 28) 0.999 999 999 273 882 091 52 × 2 = 1 + 0.999 999 998 547 764 183 04;
  • 29) 0.999 999 998 547 764 183 04 × 2 = 1 + 0.999 999 997 095 528 366 08;
  • 30) 0.999 999 997 095 528 366 08 × 2 = 1 + 0.999 999 994 191 056 732 16;
  • 31) 0.999 999 994 191 056 732 16 × 2 = 1 + 0.999 999 988 382 113 464 32;
  • 32) 0.999 999 988 382 113 464 32 × 2 = 1 + 0.999 999 976 764 226 928 64;
  • 33) 0.999 999 976 764 226 928 64 × 2 = 1 + 0.999 999 953 528 453 857 28;
  • 34) 0.999 999 953 528 453 857 28 × 2 = 1 + 0.999 999 907 056 907 714 56;
  • 35) 0.999 999 907 056 907 714 56 × 2 = 1 + 0.999 999 814 113 815 429 12;
  • 36) 0.999 999 814 113 815 429 12 × 2 = 1 + 0.999 999 628 227 630 858 24;
  • 37) 0.999 999 628 227 630 858 24 × 2 = 1 + 0.999 999 256 455 261 716 48;
  • 38) 0.999 999 256 455 261 716 48 × 2 = 1 + 0.999 998 512 910 523 432 96;
  • 39) 0.999 998 512 910 523 432 96 × 2 = 1 + 0.999 997 025 821 046 865 92;
  • 40) 0.999 997 025 821 046 865 92 × 2 = 1 + 0.999 994 051 642 093 731 84;
  • 41) 0.999 994 051 642 093 731 84 × 2 = 1 + 0.999 988 103 284 187 463 68;
  • 42) 0.999 988 103 284 187 463 68 × 2 = 1 + 0.999 976 206 568 374 927 36;
  • 43) 0.999 976 206 568 374 927 36 × 2 = 1 + 0.999 952 413 136 749 854 72;
  • 44) 0.999 952 413 136 749 854 72 × 2 = 1 + 0.999 904 826 273 499 709 44;
  • 45) 0.999 904 826 273 499 709 44 × 2 = 1 + 0.999 809 652 546 999 418 88;
  • 46) 0.999 809 652 546 999 418 88 × 2 = 1 + 0.999 619 305 093 998 837 76;
  • 47) 0.999 619 305 093 998 837 76 × 2 = 1 + 0.999 238 610 187 997 675 52;
  • 48) 0.999 238 610 187 997 675 52 × 2 = 1 + 0.998 477 220 375 995 351 04;
  • 49) 0.998 477 220 375 995 351 04 × 2 = 1 + 0.996 954 440 751 990 702 08;
  • 50) 0.996 954 440 751 990 702 08 × 2 = 1 + 0.993 908 881 503 981 404 16;
  • 51) 0.993 908 881 503 981 404 16 × 2 = 1 + 0.987 817 763 007 962 808 32;
  • 52) 0.987 817 763 007 962 808 32 × 2 = 1 + 0.975 635 526 015 925 616 64;
  • 53) 0.975 635 526 015 925 616 64 × 2 = 1 + 0.951 271 052 031 851 233 28;
  • 54) 0.951 271 052 031 851 233 28 × 2 = 1 + 0.902 542 104 063 702 466 56;
  • 55) 0.902 542 104 063 702 466 56 × 2 = 1 + 0.805 084 208 127 404 933 12;
  • 56) 0.805 084 208 127 404 933 12 × 2 = 1 + 0.610 168 416 254 809 866 24;
  • 57) 0.610 168 416 254 809 866 24 × 2 = 1 + 0.220 336 832 509 619 732 48;
  • 58) 0.220 336 832 509 619 732 48 × 2 = 0 + 0.440 673 665 019 239 464 96;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


5. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.016 738 891 601 562 494 59(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 494 59(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 494 59(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(2) × 2-6


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


12. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


Decimal number -0.016 738 891 601 562 494 59 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100