-0.016 738 891 601 562 496 530 553 004 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 004(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
-0.016 738 891 601 562 496 530 553 004(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:

|-0.016 738 891 601 562 496 530 553 004| = 0.016 738 891 601 562 496 530 553 004


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: 0.016 738 891 601 562 496 530 553 004.

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) 0.016 738 891 601 562 496 530 553 004 × 2 = 0 + 0.033 477 783 203 124 993 061 106 008;
  • 2) 0.033 477 783 203 124 993 061 106 008 × 2 = 0 + 0.066 955 566 406 249 986 122 212 016;
  • 3) 0.066 955 566 406 249 986 122 212 016 × 2 = 0 + 0.133 911 132 812 499 972 244 424 032;
  • 4) 0.133 911 132 812 499 972 244 424 032 × 2 = 0 + 0.267 822 265 624 999 944 488 848 064;
  • 5) 0.267 822 265 624 999 944 488 848 064 × 2 = 0 + 0.535 644 531 249 999 888 977 696 128;
  • 6) 0.535 644 531 249 999 888 977 696 128 × 2 = 1 + 0.071 289 062 499 999 777 955 392 256;
  • 7) 0.071 289 062 499 999 777 955 392 256 × 2 = 0 + 0.142 578 124 999 999 555 910 784 512;
  • 8) 0.142 578 124 999 999 555 910 784 512 × 2 = 0 + 0.285 156 249 999 999 111 821 569 024;
  • 9) 0.285 156 249 999 999 111 821 569 024 × 2 = 0 + 0.570 312 499 999 998 223 643 138 048;
  • 10) 0.570 312 499 999 998 223 643 138 048 × 2 = 1 + 0.140 624 999 999 996 447 286 276 096;
  • 11) 0.140 624 999 999 996 447 286 276 096 × 2 = 0 + 0.281 249 999 999 992 894 572 552 192;
  • 12) 0.281 249 999 999 992 894 572 552 192 × 2 = 0 + 0.562 499 999 999 985 789 145 104 384;
  • 13) 0.562 499 999 999 985 789 145 104 384 × 2 = 1 + 0.124 999 999 999 971 578 290 208 768;
  • 14) 0.124 999 999 999 971 578 290 208 768 × 2 = 0 + 0.249 999 999 999 943 156 580 417 536;
  • 15) 0.249 999 999 999 943 156 580 417 536 × 2 = 0 + 0.499 999 999 999 886 313 160 835 072;
  • 16) 0.499 999 999 999 886 313 160 835 072 × 2 = 0 + 0.999 999 999 999 772 626 321 670 144;
  • 17) 0.999 999 999 999 772 626 321 670 144 × 2 = 1 + 0.999 999 999 999 545 252 643 340 288;
  • 18) 0.999 999 999 999 545 252 643 340 288 × 2 = 1 + 0.999 999 999 999 090 505 286 680 576;
  • 19) 0.999 999 999 999 090 505 286 680 576 × 2 = 1 + 0.999 999 999 998 181 010 573 361 152;
  • 20) 0.999 999 999 998 181 010 573 361 152 × 2 = 1 + 0.999 999 999 996 362 021 146 722 304;
  • 21) 0.999 999 999 996 362 021 146 722 304 × 2 = 1 + 0.999 999 999 992 724 042 293 444 608;
  • 22) 0.999 999 999 992 724 042 293 444 608 × 2 = 1 + 0.999 999 999 985 448 084 586 889 216;
  • 23) 0.999 999 999 985 448 084 586 889 216 × 2 = 1 + 0.999 999 999 970 896 169 173 778 432;
  • 24) 0.999 999 999 970 896 169 173 778 432 × 2 = 1 + 0.999 999 999 941 792 338 347 556 864;
  • 25) 0.999 999 999 941 792 338 347 556 864 × 2 = 1 + 0.999 999 999 883 584 676 695 113 728;
  • 26) 0.999 999 999 883 584 676 695 113 728 × 2 = 1 + 0.999 999 999 767 169 353 390 227 456;
  • 27) 0.999 999 999 767 169 353 390 227 456 × 2 = 1 + 0.999 999 999 534 338 706 780 454 912;
  • 28) 0.999 999 999 534 338 706 780 454 912 × 2 = 1 + 0.999 999 999 068 677 413 560 909 824;
  • 29) 0.999 999 999 068 677 413 560 909 824 × 2 = 1 + 0.999 999 998 137 354 827 121 819 648;
  • 30) 0.999 999 998 137 354 827 121 819 648 × 2 = 1 + 0.999 999 996 274 709 654 243 639 296;
  • 31) 0.999 999 996 274 709 654 243 639 296 × 2 = 1 + 0.999 999 992 549 419 308 487 278 592;
  • 32) 0.999 999 992 549 419 308 487 278 592 × 2 = 1 + 0.999 999 985 098 838 616 974 557 184;
  • 33) 0.999 999 985 098 838 616 974 557 184 × 2 = 1 + 0.999 999 970 197 677 233 949 114 368;
  • 34) 0.999 999 970 197 677 233 949 114 368 × 2 = 1 + 0.999 999 940 395 354 467 898 228 736;
  • 35) 0.999 999 940 395 354 467 898 228 736 × 2 = 1 + 0.999 999 880 790 708 935 796 457 472;
  • 36) 0.999 999 880 790 708 935 796 457 472 × 2 = 1 + 0.999 999 761 581 417 871 592 914 944;
  • 37) 0.999 999 761 581 417 871 592 914 944 × 2 = 1 + 0.999 999 523 162 835 743 185 829 888;
  • 38) 0.999 999 523 162 835 743 185 829 888 × 2 = 1 + 0.999 999 046 325 671 486 371 659 776;
  • 39) 0.999 999 046 325 671 486 371 659 776 × 2 = 1 + 0.999 998 092 651 342 972 743 319 552;
  • 40) 0.999 998 092 651 342 972 743 319 552 × 2 = 1 + 0.999 996 185 302 685 945 486 639 104;
  • 41) 0.999 996 185 302 685 945 486 639 104 × 2 = 1 + 0.999 992 370 605 371 890 973 278 208;
  • 42) 0.999 992 370 605 371 890 973 278 208 × 2 = 1 + 0.999 984 741 210 743 781 946 556 416;
  • 43) 0.999 984 741 210 743 781 946 556 416 × 2 = 1 + 0.999 969 482 421 487 563 893 112 832;
  • 44) 0.999 969 482 421 487 563 893 112 832 × 2 = 1 + 0.999 938 964 842 975 127 786 225 664;
  • 45) 0.999 938 964 842 975 127 786 225 664 × 2 = 1 + 0.999 877 929 685 950 255 572 451 328;
  • 46) 0.999 877 929 685 950 255 572 451 328 × 2 = 1 + 0.999 755 859 371 900 511 144 902 656;
  • 47) 0.999 755 859 371 900 511 144 902 656 × 2 = 1 + 0.999 511 718 743 801 022 289 805 312;
  • 48) 0.999 511 718 743 801 022 289 805 312 × 2 = 1 + 0.999 023 437 487 602 044 579 610 624;
  • 49) 0.999 023 437 487 602 044 579 610 624 × 2 = 1 + 0.998 046 874 975 204 089 159 221 248;
  • 50) 0.998 046 874 975 204 089 159 221 248 × 2 = 1 + 0.996 093 749 950 408 178 318 442 496;
  • 51) 0.996 093 749 950 408 178 318 442 496 × 2 = 1 + 0.992 187 499 900 816 356 636 884 992;
  • 52) 0.992 187 499 900 816 356 636 884 992 × 2 = 1 + 0.984 374 999 801 632 713 273 769 984;
  • 53) 0.984 374 999 801 632 713 273 769 984 × 2 = 1 + 0.968 749 999 603 265 426 547 539 968;
  • 54) 0.968 749 999 603 265 426 547 539 968 × 2 = 1 + 0.937 499 999 206 530 853 095 079 936;
  • 55) 0.937 499 999 206 530 853 095 079 936 × 2 = 1 + 0.874 999 998 413 061 706 190 159 872;
  • 56) 0.874 999 998 413 061 706 190 159 872 × 2 = 1 + 0.749 999 996 826 123 412 380 319 744;
  • 57) 0.749 999 996 826 123 412 380 319 744 × 2 = 1 + 0.499 999 993 652 246 824 760 639 488;
  • 58) 0.499 999 993 652 246 824 760 639 488 × 2 = 0 + 0.999 999 987 304 493 649 521 278 976;

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:


0.016 738 891 601 562 496 530 553 004(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 004(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:


0.016 738 891 601 562 496 530 553 004(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 -0.016 738 891 601 562 496 530 553 004 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 -0.016 738 891 601 562 496 530 553 065 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