-0.016 738 891 601 562 496 530 552 957 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 552 957(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
-0.016 738 891 601 562 496 530 552 957(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:

|-0.016 738 891 601 562 496 530 552 957| = 0.016 738 891 601 562 496 530 552 957


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: 0.016 738 891 601 562 496 530 552 957.

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) 0.016 738 891 601 562 496 530 552 957 × 2 = 0 + 0.033 477 783 203 124 993 061 105 914;
  • 2) 0.033 477 783 203 124 993 061 105 914 × 2 = 0 + 0.066 955 566 406 249 986 122 211 828;
  • 3) 0.066 955 566 406 249 986 122 211 828 × 2 = 0 + 0.133 911 132 812 499 972 244 423 656;
  • 4) 0.133 911 132 812 499 972 244 423 656 × 2 = 0 + 0.267 822 265 624 999 944 488 847 312;
  • 5) 0.267 822 265 624 999 944 488 847 312 × 2 = 0 + 0.535 644 531 249 999 888 977 694 624;
  • 6) 0.535 644 531 249 999 888 977 694 624 × 2 = 1 + 0.071 289 062 499 999 777 955 389 248;
  • 7) 0.071 289 062 499 999 777 955 389 248 × 2 = 0 + 0.142 578 124 999 999 555 910 778 496;
  • 8) 0.142 578 124 999 999 555 910 778 496 × 2 = 0 + 0.285 156 249 999 999 111 821 556 992;
  • 9) 0.285 156 249 999 999 111 821 556 992 × 2 = 0 + 0.570 312 499 999 998 223 643 113 984;
  • 10) 0.570 312 499 999 998 223 643 113 984 × 2 = 1 + 0.140 624 999 999 996 447 286 227 968;
  • 11) 0.140 624 999 999 996 447 286 227 968 × 2 = 0 + 0.281 249 999 999 992 894 572 455 936;
  • 12) 0.281 249 999 999 992 894 572 455 936 × 2 = 0 + 0.562 499 999 999 985 789 144 911 872;
  • 13) 0.562 499 999 999 985 789 144 911 872 × 2 = 1 + 0.124 999 999 999 971 578 289 823 744;
  • 14) 0.124 999 999 999 971 578 289 823 744 × 2 = 0 + 0.249 999 999 999 943 156 579 647 488;
  • 15) 0.249 999 999 999 943 156 579 647 488 × 2 = 0 + 0.499 999 999 999 886 313 159 294 976;
  • 16) 0.499 999 999 999 886 313 159 294 976 × 2 = 0 + 0.999 999 999 999 772 626 318 589 952;
  • 17) 0.999 999 999 999 772 626 318 589 952 × 2 = 1 + 0.999 999 999 999 545 252 637 179 904;
  • 18) 0.999 999 999 999 545 252 637 179 904 × 2 = 1 + 0.999 999 999 999 090 505 274 359 808;
  • 19) 0.999 999 999 999 090 505 274 359 808 × 2 = 1 + 0.999 999 999 998 181 010 548 719 616;
  • 20) 0.999 999 999 998 181 010 548 719 616 × 2 = 1 + 0.999 999 999 996 362 021 097 439 232;
  • 21) 0.999 999 999 996 362 021 097 439 232 × 2 = 1 + 0.999 999 999 992 724 042 194 878 464;
  • 22) 0.999 999 999 992 724 042 194 878 464 × 2 = 1 + 0.999 999 999 985 448 084 389 756 928;
  • 23) 0.999 999 999 985 448 084 389 756 928 × 2 = 1 + 0.999 999 999 970 896 168 779 513 856;
  • 24) 0.999 999 999 970 896 168 779 513 856 × 2 = 1 + 0.999 999 999 941 792 337 559 027 712;
  • 25) 0.999 999 999 941 792 337 559 027 712 × 2 = 1 + 0.999 999 999 883 584 675 118 055 424;
  • 26) 0.999 999 999 883 584 675 118 055 424 × 2 = 1 + 0.999 999 999 767 169 350 236 110 848;
  • 27) 0.999 999 999 767 169 350 236 110 848 × 2 = 1 + 0.999 999 999 534 338 700 472 221 696;
  • 28) 0.999 999 999 534 338 700 472 221 696 × 2 = 1 + 0.999 999 999 068 677 400 944 443 392;
  • 29) 0.999 999 999 068 677 400 944 443 392 × 2 = 1 + 0.999 999 998 137 354 801 888 886 784;
  • 30) 0.999 999 998 137 354 801 888 886 784 × 2 = 1 + 0.999 999 996 274 709 603 777 773 568;
  • 31) 0.999 999 996 274 709 603 777 773 568 × 2 = 1 + 0.999 999 992 549 419 207 555 547 136;
  • 32) 0.999 999 992 549 419 207 555 547 136 × 2 = 1 + 0.999 999 985 098 838 415 111 094 272;
  • 33) 0.999 999 985 098 838 415 111 094 272 × 2 = 1 + 0.999 999 970 197 676 830 222 188 544;
  • 34) 0.999 999 970 197 676 830 222 188 544 × 2 = 1 + 0.999 999 940 395 353 660 444 377 088;
  • 35) 0.999 999 940 395 353 660 444 377 088 × 2 = 1 + 0.999 999 880 790 707 320 888 754 176;
  • 36) 0.999 999 880 790 707 320 888 754 176 × 2 = 1 + 0.999 999 761 581 414 641 777 508 352;
  • 37) 0.999 999 761 581 414 641 777 508 352 × 2 = 1 + 0.999 999 523 162 829 283 555 016 704;
  • 38) 0.999 999 523 162 829 283 555 016 704 × 2 = 1 + 0.999 999 046 325 658 567 110 033 408;
  • 39) 0.999 999 046 325 658 567 110 033 408 × 2 = 1 + 0.999 998 092 651 317 134 220 066 816;
  • 40) 0.999 998 092 651 317 134 220 066 816 × 2 = 1 + 0.999 996 185 302 634 268 440 133 632;
  • 41) 0.999 996 185 302 634 268 440 133 632 × 2 = 1 + 0.999 992 370 605 268 536 880 267 264;
  • 42) 0.999 992 370 605 268 536 880 267 264 × 2 = 1 + 0.999 984 741 210 537 073 760 534 528;
  • 43) 0.999 984 741 210 537 073 760 534 528 × 2 = 1 + 0.999 969 482 421 074 147 521 069 056;
  • 44) 0.999 969 482 421 074 147 521 069 056 × 2 = 1 + 0.999 938 964 842 148 295 042 138 112;
  • 45) 0.999 938 964 842 148 295 042 138 112 × 2 = 1 + 0.999 877 929 684 296 590 084 276 224;
  • 46) 0.999 877 929 684 296 590 084 276 224 × 2 = 1 + 0.999 755 859 368 593 180 168 552 448;
  • 47) 0.999 755 859 368 593 180 168 552 448 × 2 = 1 + 0.999 511 718 737 186 360 337 104 896;
  • 48) 0.999 511 718 737 186 360 337 104 896 × 2 = 1 + 0.999 023 437 474 372 720 674 209 792;
  • 49) 0.999 023 437 474 372 720 674 209 792 × 2 = 1 + 0.998 046 874 948 745 441 348 419 584;
  • 50) 0.998 046 874 948 745 441 348 419 584 × 2 = 1 + 0.996 093 749 897 490 882 696 839 168;
  • 51) 0.996 093 749 897 490 882 696 839 168 × 2 = 1 + 0.992 187 499 794 981 765 393 678 336;
  • 52) 0.992 187 499 794 981 765 393 678 336 × 2 = 1 + 0.984 374 999 589 963 530 787 356 672;
  • 53) 0.984 374 999 589 963 530 787 356 672 × 2 = 1 + 0.968 749 999 179 927 061 574 713 344;
  • 54) 0.968 749 999 179 927 061 574 713 344 × 2 = 1 + 0.937 499 998 359 854 123 149 426 688;
  • 55) 0.937 499 998 359 854 123 149 426 688 × 2 = 1 + 0.874 999 996 719 708 246 298 853 376;
  • 56) 0.874 999 996 719 708 246 298 853 376 × 2 = 1 + 0.749 999 993 439 416 492 597 706 752;
  • 57) 0.749 999 993 439 416 492 597 706 752 × 2 = 1 + 0.499 999 986 878 832 985 195 413 504;
  • 58) 0.499 999 986 878 832 985 195 413 504 × 2 = 0 + 0.999 999 973 757 665 970 390 827 008;

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:


0.016 738 891 601 562 496 530 552 957(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 552 957(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:


0.016 738 891 601 562 496 530 552 957(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 -0.016 738 891 601 562 496 530 552 957 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

Share

» Convert decimal -0.016 738 891 601 562 496 530 552 885 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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