-0.016 738 891 601 562 496 530 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 9(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 9(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 9| = 0.016 738 891 601 562 496 530 9


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 9.

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 9 × 2 = 0 + 0.033 477 783 203 124 993 061 8;
  • 2) 0.033 477 783 203 124 993 061 8 × 2 = 0 + 0.066 955 566 406 249 986 123 6;
  • 3) 0.066 955 566 406 249 986 123 6 × 2 = 0 + 0.133 911 132 812 499 972 247 2;
  • 4) 0.133 911 132 812 499 972 247 2 × 2 = 0 + 0.267 822 265 624 999 944 494 4;
  • 5) 0.267 822 265 624 999 944 494 4 × 2 = 0 + 0.535 644 531 249 999 888 988 8;
  • 6) 0.535 644 531 249 999 888 988 8 × 2 = 1 + 0.071 289 062 499 999 777 977 6;
  • 7) 0.071 289 062 499 999 777 977 6 × 2 = 0 + 0.142 578 124 999 999 555 955 2;
  • 8) 0.142 578 124 999 999 555 955 2 × 2 = 0 + 0.285 156 249 999 999 111 910 4;
  • 9) 0.285 156 249 999 999 111 910 4 × 2 = 0 + 0.570 312 499 999 998 223 820 8;
  • 10) 0.570 312 499 999 998 223 820 8 × 2 = 1 + 0.140 624 999 999 996 447 641 6;
  • 11) 0.140 624 999 999 996 447 641 6 × 2 = 0 + 0.281 249 999 999 992 895 283 2;
  • 12) 0.281 249 999 999 992 895 283 2 × 2 = 0 + 0.562 499 999 999 985 790 566 4;
  • 13) 0.562 499 999 999 985 790 566 4 × 2 = 1 + 0.124 999 999 999 971 581 132 8;
  • 14) 0.124 999 999 999 971 581 132 8 × 2 = 0 + 0.249 999 999 999 943 162 265 6;
  • 15) 0.249 999 999 999 943 162 265 6 × 2 = 0 + 0.499 999 999 999 886 324 531 2;
  • 16) 0.499 999 999 999 886 324 531 2 × 2 = 0 + 0.999 999 999 999 772 649 062 4;
  • 17) 0.999 999 999 999 772 649 062 4 × 2 = 1 + 0.999 999 999 999 545 298 124 8;
  • 18) 0.999 999 999 999 545 298 124 8 × 2 = 1 + 0.999 999 999 999 090 596 249 6;
  • 19) 0.999 999 999 999 090 596 249 6 × 2 = 1 + 0.999 999 999 998 181 192 499 2;
  • 20) 0.999 999 999 998 181 192 499 2 × 2 = 1 + 0.999 999 999 996 362 384 998 4;
  • 21) 0.999 999 999 996 362 384 998 4 × 2 = 1 + 0.999 999 999 992 724 769 996 8;
  • 22) 0.999 999 999 992 724 769 996 8 × 2 = 1 + 0.999 999 999 985 449 539 993 6;
  • 23) 0.999 999 999 985 449 539 993 6 × 2 = 1 + 0.999 999 999 970 899 079 987 2;
  • 24) 0.999 999 999 970 899 079 987 2 × 2 = 1 + 0.999 999 999 941 798 159 974 4;
  • 25) 0.999 999 999 941 798 159 974 4 × 2 = 1 + 0.999 999 999 883 596 319 948 8;
  • 26) 0.999 999 999 883 596 319 948 8 × 2 = 1 + 0.999 999 999 767 192 639 897 6;
  • 27) 0.999 999 999 767 192 639 897 6 × 2 = 1 + 0.999 999 999 534 385 279 795 2;
  • 28) 0.999 999 999 534 385 279 795 2 × 2 = 1 + 0.999 999 999 068 770 559 590 4;
  • 29) 0.999 999 999 068 770 559 590 4 × 2 = 1 + 0.999 999 998 137 541 119 180 8;
  • 30) 0.999 999 998 137 541 119 180 8 × 2 = 1 + 0.999 999 996 275 082 238 361 6;
  • 31) 0.999 999 996 275 082 238 361 6 × 2 = 1 + 0.999 999 992 550 164 476 723 2;
  • 32) 0.999 999 992 550 164 476 723 2 × 2 = 1 + 0.999 999 985 100 328 953 446 4;
  • 33) 0.999 999 985 100 328 953 446 4 × 2 = 1 + 0.999 999 970 200 657 906 892 8;
  • 34) 0.999 999 970 200 657 906 892 8 × 2 = 1 + 0.999 999 940 401 315 813 785 6;
  • 35) 0.999 999 940 401 315 813 785 6 × 2 = 1 + 0.999 999 880 802 631 627 571 2;
  • 36) 0.999 999 880 802 631 627 571 2 × 2 = 1 + 0.999 999 761 605 263 255 142 4;
  • 37) 0.999 999 761 605 263 255 142 4 × 2 = 1 + 0.999 999 523 210 526 510 284 8;
  • 38) 0.999 999 523 210 526 510 284 8 × 2 = 1 + 0.999 999 046 421 053 020 569 6;
  • 39) 0.999 999 046 421 053 020 569 6 × 2 = 1 + 0.999 998 092 842 106 041 139 2;
  • 40) 0.999 998 092 842 106 041 139 2 × 2 = 1 + 0.999 996 185 684 212 082 278 4;
  • 41) 0.999 996 185 684 212 082 278 4 × 2 = 1 + 0.999 992 371 368 424 164 556 8;
  • 42) 0.999 992 371 368 424 164 556 8 × 2 = 1 + 0.999 984 742 736 848 329 113 6;
  • 43) 0.999 984 742 736 848 329 113 6 × 2 = 1 + 0.999 969 485 473 696 658 227 2;
  • 44) 0.999 969 485 473 696 658 227 2 × 2 = 1 + 0.999 938 970 947 393 316 454 4;
  • 45) 0.999 938 970 947 393 316 454 4 × 2 = 1 + 0.999 877 941 894 786 632 908 8;
  • 46) 0.999 877 941 894 786 632 908 8 × 2 = 1 + 0.999 755 883 789 573 265 817 6;
  • 47) 0.999 755 883 789 573 265 817 6 × 2 = 1 + 0.999 511 767 579 146 531 635 2;
  • 48) 0.999 511 767 579 146 531 635 2 × 2 = 1 + 0.999 023 535 158 293 063 270 4;
  • 49) 0.999 023 535 158 293 063 270 4 × 2 = 1 + 0.998 047 070 316 586 126 540 8;
  • 50) 0.998 047 070 316 586 126 540 8 × 2 = 1 + 0.996 094 140 633 172 253 081 6;
  • 51) 0.996 094 140 633 172 253 081 6 × 2 = 1 + 0.992 188 281 266 344 506 163 2;
  • 52) 0.992 188 281 266 344 506 163 2 × 2 = 1 + 0.984 376 562 532 689 012 326 4;
  • 53) 0.984 376 562 532 689 012 326 4 × 2 = 1 + 0.968 753 125 065 378 024 652 8;
  • 54) 0.968 753 125 065 378 024 652 8 × 2 = 1 + 0.937 506 250 130 756 049 305 6;
  • 55) 0.937 506 250 130 756 049 305 6 × 2 = 1 + 0.875 012 500 261 512 098 611 2;
  • 56) 0.875 012 500 261 512 098 611 2 × 2 = 1 + 0.750 025 000 523 024 197 222 4;
  • 57) 0.750 025 000 523 024 197 222 4 × 2 = 1 + 0.500 050 001 046 048 394 444 8;
  • 58) 0.500 050 001 046 048 394 444 8 × 2 = 1 + 0.000 100 002 092 096 788 889 6;

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 9(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 9(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:


0.016 738 891 601 562 496 530 9(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 -0.016 738 891 601 562 496 530 9 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

Share

» Convert decimal -0.016 738 891 601 562 496 521 4 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 06, 2026 [June]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: June - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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