-0.016 738 891 601 562 496 529 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 529 35(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
-0.016 738 891 601 562 496 529 35(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:

|-0.016 738 891 601 562 496 529 35| = 0.016 738 891 601 562 496 529 35


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: 0.016 738 891 601 562 496 529 35.

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) 0.016 738 891 601 562 496 529 35 × 2 = 0 + 0.033 477 783 203 124 993 058 7;
  • 2) 0.033 477 783 203 124 993 058 7 × 2 = 0 + 0.066 955 566 406 249 986 117 4;
  • 3) 0.066 955 566 406 249 986 117 4 × 2 = 0 + 0.133 911 132 812 499 972 234 8;
  • 4) 0.133 911 132 812 499 972 234 8 × 2 = 0 + 0.267 822 265 624 999 944 469 6;
  • 5) 0.267 822 265 624 999 944 469 6 × 2 = 0 + 0.535 644 531 249 999 888 939 2;
  • 6) 0.535 644 531 249 999 888 939 2 × 2 = 1 + 0.071 289 062 499 999 777 878 4;
  • 7) 0.071 289 062 499 999 777 878 4 × 2 = 0 + 0.142 578 124 999 999 555 756 8;
  • 8) 0.142 578 124 999 999 555 756 8 × 2 = 0 + 0.285 156 249 999 999 111 513 6;
  • 9) 0.285 156 249 999 999 111 513 6 × 2 = 0 + 0.570 312 499 999 998 223 027 2;
  • 10) 0.570 312 499 999 998 223 027 2 × 2 = 1 + 0.140 624 999 999 996 446 054 4;
  • 11) 0.140 624 999 999 996 446 054 4 × 2 = 0 + 0.281 249 999 999 992 892 108 8;
  • 12) 0.281 249 999 999 992 892 108 8 × 2 = 0 + 0.562 499 999 999 985 784 217 6;
  • 13) 0.562 499 999 999 985 784 217 6 × 2 = 1 + 0.124 999 999 999 971 568 435 2;
  • 14) 0.124 999 999 999 971 568 435 2 × 2 = 0 + 0.249 999 999 999 943 136 870 4;
  • 15) 0.249 999 999 999 943 136 870 4 × 2 = 0 + 0.499 999 999 999 886 273 740 8;
  • 16) 0.499 999 999 999 886 273 740 8 × 2 = 0 + 0.999 999 999 999 772 547 481 6;
  • 17) 0.999 999 999 999 772 547 481 6 × 2 = 1 + 0.999 999 999 999 545 094 963 2;
  • 18) 0.999 999 999 999 545 094 963 2 × 2 = 1 + 0.999 999 999 999 090 189 926 4;
  • 19) 0.999 999 999 999 090 189 926 4 × 2 = 1 + 0.999 999 999 998 180 379 852 8;
  • 20) 0.999 999 999 998 180 379 852 8 × 2 = 1 + 0.999 999 999 996 360 759 705 6;
  • 21) 0.999 999 999 996 360 759 705 6 × 2 = 1 + 0.999 999 999 992 721 519 411 2;
  • 22) 0.999 999 999 992 721 519 411 2 × 2 = 1 + 0.999 999 999 985 443 038 822 4;
  • 23) 0.999 999 999 985 443 038 822 4 × 2 = 1 + 0.999 999 999 970 886 077 644 8;
  • 24) 0.999 999 999 970 886 077 644 8 × 2 = 1 + 0.999 999 999 941 772 155 289 6;
  • 25) 0.999 999 999 941 772 155 289 6 × 2 = 1 + 0.999 999 999 883 544 310 579 2;
  • 26) 0.999 999 999 883 544 310 579 2 × 2 = 1 + 0.999 999 999 767 088 621 158 4;
  • 27) 0.999 999 999 767 088 621 158 4 × 2 = 1 + 0.999 999 999 534 177 242 316 8;
  • 28) 0.999 999 999 534 177 242 316 8 × 2 = 1 + 0.999 999 999 068 354 484 633 6;
  • 29) 0.999 999 999 068 354 484 633 6 × 2 = 1 + 0.999 999 998 136 708 969 267 2;
  • 30) 0.999 999 998 136 708 969 267 2 × 2 = 1 + 0.999 999 996 273 417 938 534 4;
  • 31) 0.999 999 996 273 417 938 534 4 × 2 = 1 + 0.999 999 992 546 835 877 068 8;
  • 32) 0.999 999 992 546 835 877 068 8 × 2 = 1 + 0.999 999 985 093 671 754 137 6;
  • 33) 0.999 999 985 093 671 754 137 6 × 2 = 1 + 0.999 999 970 187 343 508 275 2;
  • 34) 0.999 999 970 187 343 508 275 2 × 2 = 1 + 0.999 999 940 374 687 016 550 4;
  • 35) 0.999 999 940 374 687 016 550 4 × 2 = 1 + 0.999 999 880 749 374 033 100 8;
  • 36) 0.999 999 880 749 374 033 100 8 × 2 = 1 + 0.999 999 761 498 748 066 201 6;
  • 37) 0.999 999 761 498 748 066 201 6 × 2 = 1 + 0.999 999 522 997 496 132 403 2;
  • 38) 0.999 999 522 997 496 132 403 2 × 2 = 1 + 0.999 999 045 994 992 264 806 4;
  • 39) 0.999 999 045 994 992 264 806 4 × 2 = 1 + 0.999 998 091 989 984 529 612 8;
  • 40) 0.999 998 091 989 984 529 612 8 × 2 = 1 + 0.999 996 183 979 969 059 225 6;
  • 41) 0.999 996 183 979 969 059 225 6 × 2 = 1 + 0.999 992 367 959 938 118 451 2;
  • 42) 0.999 992 367 959 938 118 451 2 × 2 = 1 + 0.999 984 735 919 876 236 902 4;
  • 43) 0.999 984 735 919 876 236 902 4 × 2 = 1 + 0.999 969 471 839 752 473 804 8;
  • 44) 0.999 969 471 839 752 473 804 8 × 2 = 1 + 0.999 938 943 679 504 947 609 6;
  • 45) 0.999 938 943 679 504 947 609 6 × 2 = 1 + 0.999 877 887 359 009 895 219 2;
  • 46) 0.999 877 887 359 009 895 219 2 × 2 = 1 + 0.999 755 774 718 019 790 438 4;
  • 47) 0.999 755 774 718 019 790 438 4 × 2 = 1 + 0.999 511 549 436 039 580 876 8;
  • 48) 0.999 511 549 436 039 580 876 8 × 2 = 1 + 0.999 023 098 872 079 161 753 6;
  • 49) 0.999 023 098 872 079 161 753 6 × 2 = 1 + 0.998 046 197 744 158 323 507 2;
  • 50) 0.998 046 197 744 158 323 507 2 × 2 = 1 + 0.996 092 395 488 316 647 014 4;
  • 51) 0.996 092 395 488 316 647 014 4 × 2 = 1 + 0.992 184 790 976 633 294 028 8;
  • 52) 0.992 184 790 976 633 294 028 8 × 2 = 1 + 0.984 369 581 953 266 588 057 6;
  • 53) 0.984 369 581 953 266 588 057 6 × 2 = 1 + 0.968 739 163 906 533 176 115 2;
  • 54) 0.968 739 163 906 533 176 115 2 × 2 = 1 + 0.937 478 327 813 066 352 230 4;
  • 55) 0.937 478 327 813 066 352 230 4 × 2 = 1 + 0.874 956 655 626 132 704 460 8;
  • 56) 0.874 956 655 626 132 704 460 8 × 2 = 1 + 0.749 913 311 252 265 408 921 6;
  • 57) 0.749 913 311 252 265 408 921 6 × 2 = 1 + 0.499 826 622 504 530 817 843 2;
  • 58) 0.499 826 622 504 530 817 843 2 × 2 = 0 + 0.999 653 245 009 061 635 686 4;

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:


0.016 738 891 601 562 496 529 35(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 529 35(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:


0.016 738 891 601 562 496 529 35(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 -0.016 738 891 601 562 496 529 35 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