-0.016 738 891 601 562 496 530 553 046 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 046 3(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
-0.016 738 891 601 562 496 530 553 046 3(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:

|-0.016 738 891 601 562 496 530 553 046 3| = 0.016 738 891 601 562 496 530 553 046 3


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: 0.016 738 891 601 562 496 530 553 046 3.

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) 0.016 738 891 601 562 496 530 553 046 3 × 2 = 0 + 0.033 477 783 203 124 993 061 106 092 6;
  • 2) 0.033 477 783 203 124 993 061 106 092 6 × 2 = 0 + 0.066 955 566 406 249 986 122 212 185 2;
  • 3) 0.066 955 566 406 249 986 122 212 185 2 × 2 = 0 + 0.133 911 132 812 499 972 244 424 370 4;
  • 4) 0.133 911 132 812 499 972 244 424 370 4 × 2 = 0 + 0.267 822 265 624 999 944 488 848 740 8;
  • 5) 0.267 822 265 624 999 944 488 848 740 8 × 2 = 0 + 0.535 644 531 249 999 888 977 697 481 6;
  • 6) 0.535 644 531 249 999 888 977 697 481 6 × 2 = 1 + 0.071 289 062 499 999 777 955 394 963 2;
  • 7) 0.071 289 062 499 999 777 955 394 963 2 × 2 = 0 + 0.142 578 124 999 999 555 910 789 926 4;
  • 8) 0.142 578 124 999 999 555 910 789 926 4 × 2 = 0 + 0.285 156 249 999 999 111 821 579 852 8;
  • 9) 0.285 156 249 999 999 111 821 579 852 8 × 2 = 0 + 0.570 312 499 999 998 223 643 159 705 6;
  • 10) 0.570 312 499 999 998 223 643 159 705 6 × 2 = 1 + 0.140 624 999 999 996 447 286 319 411 2;
  • 11) 0.140 624 999 999 996 447 286 319 411 2 × 2 = 0 + 0.281 249 999 999 992 894 572 638 822 4;
  • 12) 0.281 249 999 999 992 894 572 638 822 4 × 2 = 0 + 0.562 499 999 999 985 789 145 277 644 8;
  • 13) 0.562 499 999 999 985 789 145 277 644 8 × 2 = 1 + 0.124 999 999 999 971 578 290 555 289 6;
  • 14) 0.124 999 999 999 971 578 290 555 289 6 × 2 = 0 + 0.249 999 999 999 943 156 581 110 579 2;
  • 15) 0.249 999 999 999 943 156 581 110 579 2 × 2 = 0 + 0.499 999 999 999 886 313 162 221 158 4;
  • 16) 0.499 999 999 999 886 313 162 221 158 4 × 2 = 0 + 0.999 999 999 999 772 626 324 442 316 8;
  • 17) 0.999 999 999 999 772 626 324 442 316 8 × 2 = 1 + 0.999 999 999 999 545 252 648 884 633 6;
  • 18) 0.999 999 999 999 545 252 648 884 633 6 × 2 = 1 + 0.999 999 999 999 090 505 297 769 267 2;
  • 19) 0.999 999 999 999 090 505 297 769 267 2 × 2 = 1 + 0.999 999 999 998 181 010 595 538 534 4;
  • 20) 0.999 999 999 998 181 010 595 538 534 4 × 2 = 1 + 0.999 999 999 996 362 021 191 077 068 8;
  • 21) 0.999 999 999 996 362 021 191 077 068 8 × 2 = 1 + 0.999 999 999 992 724 042 382 154 137 6;
  • 22) 0.999 999 999 992 724 042 382 154 137 6 × 2 = 1 + 0.999 999 999 985 448 084 764 308 275 2;
  • 23) 0.999 999 999 985 448 084 764 308 275 2 × 2 = 1 + 0.999 999 999 970 896 169 528 616 550 4;
  • 24) 0.999 999 999 970 896 169 528 616 550 4 × 2 = 1 + 0.999 999 999 941 792 339 057 233 100 8;
  • 25) 0.999 999 999 941 792 339 057 233 100 8 × 2 = 1 + 0.999 999 999 883 584 678 114 466 201 6;
  • 26) 0.999 999 999 883 584 678 114 466 201 6 × 2 = 1 + 0.999 999 999 767 169 356 228 932 403 2;
  • 27) 0.999 999 999 767 169 356 228 932 403 2 × 2 = 1 + 0.999 999 999 534 338 712 457 864 806 4;
  • 28) 0.999 999 999 534 338 712 457 864 806 4 × 2 = 1 + 0.999 999 999 068 677 424 915 729 612 8;
  • 29) 0.999 999 999 068 677 424 915 729 612 8 × 2 = 1 + 0.999 999 998 137 354 849 831 459 225 6;
  • 30) 0.999 999 998 137 354 849 831 459 225 6 × 2 = 1 + 0.999 999 996 274 709 699 662 918 451 2;
  • 31) 0.999 999 996 274 709 699 662 918 451 2 × 2 = 1 + 0.999 999 992 549 419 399 325 836 902 4;
  • 32) 0.999 999 992 549 419 399 325 836 902 4 × 2 = 1 + 0.999 999 985 098 838 798 651 673 804 8;
  • 33) 0.999 999 985 098 838 798 651 673 804 8 × 2 = 1 + 0.999 999 970 197 677 597 303 347 609 6;
  • 34) 0.999 999 970 197 677 597 303 347 609 6 × 2 = 1 + 0.999 999 940 395 355 194 606 695 219 2;
  • 35) 0.999 999 940 395 355 194 606 695 219 2 × 2 = 1 + 0.999 999 880 790 710 389 213 390 438 4;
  • 36) 0.999 999 880 790 710 389 213 390 438 4 × 2 = 1 + 0.999 999 761 581 420 778 426 780 876 8;
  • 37) 0.999 999 761 581 420 778 426 780 876 8 × 2 = 1 + 0.999 999 523 162 841 556 853 561 753 6;
  • 38) 0.999 999 523 162 841 556 853 561 753 6 × 2 = 1 + 0.999 999 046 325 683 113 707 123 507 2;
  • 39) 0.999 999 046 325 683 113 707 123 507 2 × 2 = 1 + 0.999 998 092 651 366 227 414 247 014 4;
  • 40) 0.999 998 092 651 366 227 414 247 014 4 × 2 = 1 + 0.999 996 185 302 732 454 828 494 028 8;
  • 41) 0.999 996 185 302 732 454 828 494 028 8 × 2 = 1 + 0.999 992 370 605 464 909 656 988 057 6;
  • 42) 0.999 992 370 605 464 909 656 988 057 6 × 2 = 1 + 0.999 984 741 210 929 819 313 976 115 2;
  • 43) 0.999 984 741 210 929 819 313 976 115 2 × 2 = 1 + 0.999 969 482 421 859 638 627 952 230 4;
  • 44) 0.999 969 482 421 859 638 627 952 230 4 × 2 = 1 + 0.999 938 964 843 719 277 255 904 460 8;
  • 45) 0.999 938 964 843 719 277 255 904 460 8 × 2 = 1 + 0.999 877 929 687 438 554 511 808 921 6;
  • 46) 0.999 877 929 687 438 554 511 808 921 6 × 2 = 1 + 0.999 755 859 374 877 109 023 617 843 2;
  • 47) 0.999 755 859 374 877 109 023 617 843 2 × 2 = 1 + 0.999 511 718 749 754 218 047 235 686 4;
  • 48) 0.999 511 718 749 754 218 047 235 686 4 × 2 = 1 + 0.999 023 437 499 508 436 094 471 372 8;
  • 49) 0.999 023 437 499 508 436 094 471 372 8 × 2 = 1 + 0.998 046 874 999 016 872 188 942 745 6;
  • 50) 0.998 046 874 999 016 872 188 942 745 6 × 2 = 1 + 0.996 093 749 998 033 744 377 885 491 2;
  • 51) 0.996 093 749 998 033 744 377 885 491 2 × 2 = 1 + 0.992 187 499 996 067 488 755 770 982 4;
  • 52) 0.992 187 499 996 067 488 755 770 982 4 × 2 = 1 + 0.984 374 999 992 134 977 511 541 964 8;
  • 53) 0.984 374 999 992 134 977 511 541 964 8 × 2 = 1 + 0.968 749 999 984 269 955 023 083 929 6;
  • 54) 0.968 749 999 984 269 955 023 083 929 6 × 2 = 1 + 0.937 499 999 968 539 910 046 167 859 2;
  • 55) 0.937 499 999 968 539 910 046 167 859 2 × 2 = 1 + 0.874 999 999 937 079 820 092 335 718 4;
  • 56) 0.874 999 999 937 079 820 092 335 718 4 × 2 = 1 + 0.749 999 999 874 159 640 184 671 436 8;
  • 57) 0.749 999 999 874 159 640 184 671 436 8 × 2 = 1 + 0.499 999 999 748 319 280 369 342 873 6;
  • 58) 0.499 999 999 748 319 280 369 342 873 6 × 2 = 0 + 0.999 999 999 496 638 560 738 685 747 2;

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:


0.016 738 891 601 562 496 530 553 046 3(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 046 3(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:


0.016 738 891 601 562 496 530 553 046 3(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 -0.016 738 891 601 562 496 530 553 046 3 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 -0.016 738 891 601 562 496 530 553 047 5 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