-0.016 738 891 601 562 496 530 366 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 366(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
-0.016 738 891 601 562 496 530 366(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:

|-0.016 738 891 601 562 496 530 366| = 0.016 738 891 601 562 496 530 366


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: 0.016 738 891 601 562 496 530 366.

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) 0.016 738 891 601 562 496 530 366 × 2 = 0 + 0.033 477 783 203 124 993 060 732;
  • 2) 0.033 477 783 203 124 993 060 732 × 2 = 0 + 0.066 955 566 406 249 986 121 464;
  • 3) 0.066 955 566 406 249 986 121 464 × 2 = 0 + 0.133 911 132 812 499 972 242 928;
  • 4) 0.133 911 132 812 499 972 242 928 × 2 = 0 + 0.267 822 265 624 999 944 485 856;
  • 5) 0.267 822 265 624 999 944 485 856 × 2 = 0 + 0.535 644 531 249 999 888 971 712;
  • 6) 0.535 644 531 249 999 888 971 712 × 2 = 1 + 0.071 289 062 499 999 777 943 424;
  • 7) 0.071 289 062 499 999 777 943 424 × 2 = 0 + 0.142 578 124 999 999 555 886 848;
  • 8) 0.142 578 124 999 999 555 886 848 × 2 = 0 + 0.285 156 249 999 999 111 773 696;
  • 9) 0.285 156 249 999 999 111 773 696 × 2 = 0 + 0.570 312 499 999 998 223 547 392;
  • 10) 0.570 312 499 999 998 223 547 392 × 2 = 1 + 0.140 624 999 999 996 447 094 784;
  • 11) 0.140 624 999 999 996 447 094 784 × 2 = 0 + 0.281 249 999 999 992 894 189 568;
  • 12) 0.281 249 999 999 992 894 189 568 × 2 = 0 + 0.562 499 999 999 985 788 379 136;
  • 13) 0.562 499 999 999 985 788 379 136 × 2 = 1 + 0.124 999 999 999 971 576 758 272;
  • 14) 0.124 999 999 999 971 576 758 272 × 2 = 0 + 0.249 999 999 999 943 153 516 544;
  • 15) 0.249 999 999 999 943 153 516 544 × 2 = 0 + 0.499 999 999 999 886 307 033 088;
  • 16) 0.499 999 999 999 886 307 033 088 × 2 = 0 + 0.999 999 999 999 772 614 066 176;
  • 17) 0.999 999 999 999 772 614 066 176 × 2 = 1 + 0.999 999 999 999 545 228 132 352;
  • 18) 0.999 999 999 999 545 228 132 352 × 2 = 1 + 0.999 999 999 999 090 456 264 704;
  • 19) 0.999 999 999 999 090 456 264 704 × 2 = 1 + 0.999 999 999 998 180 912 529 408;
  • 20) 0.999 999 999 998 180 912 529 408 × 2 = 1 + 0.999 999 999 996 361 825 058 816;
  • 21) 0.999 999 999 996 361 825 058 816 × 2 = 1 + 0.999 999 999 992 723 650 117 632;
  • 22) 0.999 999 999 992 723 650 117 632 × 2 = 1 + 0.999 999 999 985 447 300 235 264;
  • 23) 0.999 999 999 985 447 300 235 264 × 2 = 1 + 0.999 999 999 970 894 600 470 528;
  • 24) 0.999 999 999 970 894 600 470 528 × 2 = 1 + 0.999 999 999 941 789 200 941 056;
  • 25) 0.999 999 999 941 789 200 941 056 × 2 = 1 + 0.999 999 999 883 578 401 882 112;
  • 26) 0.999 999 999 883 578 401 882 112 × 2 = 1 + 0.999 999 999 767 156 803 764 224;
  • 27) 0.999 999 999 767 156 803 764 224 × 2 = 1 + 0.999 999 999 534 313 607 528 448;
  • 28) 0.999 999 999 534 313 607 528 448 × 2 = 1 + 0.999 999 999 068 627 215 056 896;
  • 29) 0.999 999 999 068 627 215 056 896 × 2 = 1 + 0.999 999 998 137 254 430 113 792;
  • 30) 0.999 999 998 137 254 430 113 792 × 2 = 1 + 0.999 999 996 274 508 860 227 584;
  • 31) 0.999 999 996 274 508 860 227 584 × 2 = 1 + 0.999 999 992 549 017 720 455 168;
  • 32) 0.999 999 992 549 017 720 455 168 × 2 = 1 + 0.999 999 985 098 035 440 910 336;
  • 33) 0.999 999 985 098 035 440 910 336 × 2 = 1 + 0.999 999 970 196 070 881 820 672;
  • 34) 0.999 999 970 196 070 881 820 672 × 2 = 1 + 0.999 999 940 392 141 763 641 344;
  • 35) 0.999 999 940 392 141 763 641 344 × 2 = 1 + 0.999 999 880 784 283 527 282 688;
  • 36) 0.999 999 880 784 283 527 282 688 × 2 = 1 + 0.999 999 761 568 567 054 565 376;
  • 37) 0.999 999 761 568 567 054 565 376 × 2 = 1 + 0.999 999 523 137 134 109 130 752;
  • 38) 0.999 999 523 137 134 109 130 752 × 2 = 1 + 0.999 999 046 274 268 218 261 504;
  • 39) 0.999 999 046 274 268 218 261 504 × 2 = 1 + 0.999 998 092 548 536 436 523 008;
  • 40) 0.999 998 092 548 536 436 523 008 × 2 = 1 + 0.999 996 185 097 072 873 046 016;
  • 41) 0.999 996 185 097 072 873 046 016 × 2 = 1 + 0.999 992 370 194 145 746 092 032;
  • 42) 0.999 992 370 194 145 746 092 032 × 2 = 1 + 0.999 984 740 388 291 492 184 064;
  • 43) 0.999 984 740 388 291 492 184 064 × 2 = 1 + 0.999 969 480 776 582 984 368 128;
  • 44) 0.999 969 480 776 582 984 368 128 × 2 = 1 + 0.999 938 961 553 165 968 736 256;
  • 45) 0.999 938 961 553 165 968 736 256 × 2 = 1 + 0.999 877 923 106 331 937 472 512;
  • 46) 0.999 877 923 106 331 937 472 512 × 2 = 1 + 0.999 755 846 212 663 874 945 024;
  • 47) 0.999 755 846 212 663 874 945 024 × 2 = 1 + 0.999 511 692 425 327 749 890 048;
  • 48) 0.999 511 692 425 327 749 890 048 × 2 = 1 + 0.999 023 384 850 655 499 780 096;
  • 49) 0.999 023 384 850 655 499 780 096 × 2 = 1 + 0.998 046 769 701 310 999 560 192;
  • 50) 0.998 046 769 701 310 999 560 192 × 2 = 1 + 0.996 093 539 402 621 999 120 384;
  • 51) 0.996 093 539 402 621 999 120 384 × 2 = 1 + 0.992 187 078 805 243 998 240 768;
  • 52) 0.992 187 078 805 243 998 240 768 × 2 = 1 + 0.984 374 157 610 487 996 481 536;
  • 53) 0.984 374 157 610 487 996 481 536 × 2 = 1 + 0.968 748 315 220 975 992 963 072;
  • 54) 0.968 748 315 220 975 992 963 072 × 2 = 1 + 0.937 496 630 441 951 985 926 144;
  • 55) 0.937 496 630 441 951 985 926 144 × 2 = 1 + 0.874 993 260 883 903 971 852 288;
  • 56) 0.874 993 260 883 903 971 852 288 × 2 = 1 + 0.749 986 521 767 807 943 704 576;
  • 57) 0.749 986 521 767 807 943 704 576 × 2 = 1 + 0.499 973 043 535 615 887 409 152;
  • 58) 0.499 973 043 535 615 887 409 152 × 2 = 0 + 0.999 946 087 071 231 774 818 304;

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:


0.016 738 891 601 562 496 530 366(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 366(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:


0.016 738 891 601 562 496 530 366(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 -0.016 738 891 601 562 496 530 366 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