-0.016 738 891 601 562 496 944 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 944(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
-0.016 738 891 601 562 496 944(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:

|-0.016 738 891 601 562 496 944| = 0.016 738 891 601 562 496 944


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: 0.016 738 891 601 562 496 944.

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) 0.016 738 891 601 562 496 944 × 2 = 0 + 0.033 477 783 203 124 993 888;
  • 2) 0.033 477 783 203 124 993 888 × 2 = 0 + 0.066 955 566 406 249 987 776;
  • 3) 0.066 955 566 406 249 987 776 × 2 = 0 + 0.133 911 132 812 499 975 552;
  • 4) 0.133 911 132 812 499 975 552 × 2 = 0 + 0.267 822 265 624 999 951 104;
  • 5) 0.267 822 265 624 999 951 104 × 2 = 0 + 0.535 644 531 249 999 902 208;
  • 6) 0.535 644 531 249 999 902 208 × 2 = 1 + 0.071 289 062 499 999 804 416;
  • 7) 0.071 289 062 499 999 804 416 × 2 = 0 + 0.142 578 124 999 999 608 832;
  • 8) 0.142 578 124 999 999 608 832 × 2 = 0 + 0.285 156 249 999 999 217 664;
  • 9) 0.285 156 249 999 999 217 664 × 2 = 0 + 0.570 312 499 999 998 435 328;
  • 10) 0.570 312 499 999 998 435 328 × 2 = 1 + 0.140 624 999 999 996 870 656;
  • 11) 0.140 624 999 999 996 870 656 × 2 = 0 + 0.281 249 999 999 993 741 312;
  • 12) 0.281 249 999 999 993 741 312 × 2 = 0 + 0.562 499 999 999 987 482 624;
  • 13) 0.562 499 999 999 987 482 624 × 2 = 1 + 0.124 999 999 999 974 965 248;
  • 14) 0.124 999 999 999 974 965 248 × 2 = 0 + 0.249 999 999 999 949 930 496;
  • 15) 0.249 999 999 999 949 930 496 × 2 = 0 + 0.499 999 999 999 899 860 992;
  • 16) 0.499 999 999 999 899 860 992 × 2 = 0 + 0.999 999 999 999 799 721 984;
  • 17) 0.999 999 999 999 799 721 984 × 2 = 1 + 0.999 999 999 999 599 443 968;
  • 18) 0.999 999 999 999 599 443 968 × 2 = 1 + 0.999 999 999 999 198 887 936;
  • 19) 0.999 999 999 999 198 887 936 × 2 = 1 + 0.999 999 999 998 397 775 872;
  • 20) 0.999 999 999 998 397 775 872 × 2 = 1 + 0.999 999 999 996 795 551 744;
  • 21) 0.999 999 999 996 795 551 744 × 2 = 1 + 0.999 999 999 993 591 103 488;
  • 22) 0.999 999 999 993 591 103 488 × 2 = 1 + 0.999 999 999 987 182 206 976;
  • 23) 0.999 999 999 987 182 206 976 × 2 = 1 + 0.999 999 999 974 364 413 952;
  • 24) 0.999 999 999 974 364 413 952 × 2 = 1 + 0.999 999 999 948 728 827 904;
  • 25) 0.999 999 999 948 728 827 904 × 2 = 1 + 0.999 999 999 897 457 655 808;
  • 26) 0.999 999 999 897 457 655 808 × 2 = 1 + 0.999 999 999 794 915 311 616;
  • 27) 0.999 999 999 794 915 311 616 × 2 = 1 + 0.999 999 999 589 830 623 232;
  • 28) 0.999 999 999 589 830 623 232 × 2 = 1 + 0.999 999 999 179 661 246 464;
  • 29) 0.999 999 999 179 661 246 464 × 2 = 1 + 0.999 999 998 359 322 492 928;
  • 30) 0.999 999 998 359 322 492 928 × 2 = 1 + 0.999 999 996 718 644 985 856;
  • 31) 0.999 999 996 718 644 985 856 × 2 = 1 + 0.999 999 993 437 289 971 712;
  • 32) 0.999 999 993 437 289 971 712 × 2 = 1 + 0.999 999 986 874 579 943 424;
  • 33) 0.999 999 986 874 579 943 424 × 2 = 1 + 0.999 999 973 749 159 886 848;
  • 34) 0.999 999 973 749 159 886 848 × 2 = 1 + 0.999 999 947 498 319 773 696;
  • 35) 0.999 999 947 498 319 773 696 × 2 = 1 + 0.999 999 894 996 639 547 392;
  • 36) 0.999 999 894 996 639 547 392 × 2 = 1 + 0.999 999 789 993 279 094 784;
  • 37) 0.999 999 789 993 279 094 784 × 2 = 1 + 0.999 999 579 986 558 189 568;
  • 38) 0.999 999 579 986 558 189 568 × 2 = 1 + 0.999 999 159 973 116 379 136;
  • 39) 0.999 999 159 973 116 379 136 × 2 = 1 + 0.999 998 319 946 232 758 272;
  • 40) 0.999 998 319 946 232 758 272 × 2 = 1 + 0.999 996 639 892 465 516 544;
  • 41) 0.999 996 639 892 465 516 544 × 2 = 1 + 0.999 993 279 784 931 033 088;
  • 42) 0.999 993 279 784 931 033 088 × 2 = 1 + 0.999 986 559 569 862 066 176;
  • 43) 0.999 986 559 569 862 066 176 × 2 = 1 + 0.999 973 119 139 724 132 352;
  • 44) 0.999 973 119 139 724 132 352 × 2 = 1 + 0.999 946 238 279 448 264 704;
  • 45) 0.999 946 238 279 448 264 704 × 2 = 1 + 0.999 892 476 558 896 529 408;
  • 46) 0.999 892 476 558 896 529 408 × 2 = 1 + 0.999 784 953 117 793 058 816;
  • 47) 0.999 784 953 117 793 058 816 × 2 = 1 + 0.999 569 906 235 586 117 632;
  • 48) 0.999 569 906 235 586 117 632 × 2 = 1 + 0.999 139 812 471 172 235 264;
  • 49) 0.999 139 812 471 172 235 264 × 2 = 1 + 0.998 279 624 942 344 470 528;
  • 50) 0.998 279 624 942 344 470 528 × 2 = 1 + 0.996 559 249 884 688 941 056;
  • 51) 0.996 559 249 884 688 941 056 × 2 = 1 + 0.993 118 499 769 377 882 112;
  • 52) 0.993 118 499 769 377 882 112 × 2 = 1 + 0.986 236 999 538 755 764 224;
  • 53) 0.986 236 999 538 755 764 224 × 2 = 1 + 0.972 473 999 077 511 528 448;
  • 54) 0.972 473 999 077 511 528 448 × 2 = 1 + 0.944 947 998 155 023 056 896;
  • 55) 0.944 947 998 155 023 056 896 × 2 = 1 + 0.889 895 996 310 046 113 792;
  • 56) 0.889 895 996 310 046 113 792 × 2 = 1 + 0.779 791 992 620 092 227 584;
  • 57) 0.779 791 992 620 092 227 584 × 2 = 1 + 0.559 583 985 240 184 455 168;
  • 58) 0.559 583 985 240 184 455 168 × 2 = 1 + 0.119 167 970 480 368 910 336;

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:


0.016 738 891 601 562 496 944(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 944(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(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:


0.016 738 891 601 562 496 944(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(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 1111


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 1111 =

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


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 1111


Decimal number -0.016 738 891 601 562 496 944 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 1111

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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