-0.016 738 891 601 562 496 530 553 214 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 214(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
-0.016 738 891 601 562 496 530 553 214(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:

|-0.016 738 891 601 562 496 530 553 214| = 0.016 738 891 601 562 496 530 553 214


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: 0.016 738 891 601 562 496 530 553 214.

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) 0.016 738 891 601 562 496 530 553 214 × 2 = 0 + 0.033 477 783 203 124 993 061 106 428;
  • 2) 0.033 477 783 203 124 993 061 106 428 × 2 = 0 + 0.066 955 566 406 249 986 122 212 856;
  • 3) 0.066 955 566 406 249 986 122 212 856 × 2 = 0 + 0.133 911 132 812 499 972 244 425 712;
  • 4) 0.133 911 132 812 499 972 244 425 712 × 2 = 0 + 0.267 822 265 624 999 944 488 851 424;
  • 5) 0.267 822 265 624 999 944 488 851 424 × 2 = 0 + 0.535 644 531 249 999 888 977 702 848;
  • 6) 0.535 644 531 249 999 888 977 702 848 × 2 = 1 + 0.071 289 062 499 999 777 955 405 696;
  • 7) 0.071 289 062 499 999 777 955 405 696 × 2 = 0 + 0.142 578 124 999 999 555 910 811 392;
  • 8) 0.142 578 124 999 999 555 910 811 392 × 2 = 0 + 0.285 156 249 999 999 111 821 622 784;
  • 9) 0.285 156 249 999 999 111 821 622 784 × 2 = 0 + 0.570 312 499 999 998 223 643 245 568;
  • 10) 0.570 312 499 999 998 223 643 245 568 × 2 = 1 + 0.140 624 999 999 996 447 286 491 136;
  • 11) 0.140 624 999 999 996 447 286 491 136 × 2 = 0 + 0.281 249 999 999 992 894 572 982 272;
  • 12) 0.281 249 999 999 992 894 572 982 272 × 2 = 0 + 0.562 499 999 999 985 789 145 964 544;
  • 13) 0.562 499 999 999 985 789 145 964 544 × 2 = 1 + 0.124 999 999 999 971 578 291 929 088;
  • 14) 0.124 999 999 999 971 578 291 929 088 × 2 = 0 + 0.249 999 999 999 943 156 583 858 176;
  • 15) 0.249 999 999 999 943 156 583 858 176 × 2 = 0 + 0.499 999 999 999 886 313 167 716 352;
  • 16) 0.499 999 999 999 886 313 167 716 352 × 2 = 0 + 0.999 999 999 999 772 626 335 432 704;
  • 17) 0.999 999 999 999 772 626 335 432 704 × 2 = 1 + 0.999 999 999 999 545 252 670 865 408;
  • 18) 0.999 999 999 999 545 252 670 865 408 × 2 = 1 + 0.999 999 999 999 090 505 341 730 816;
  • 19) 0.999 999 999 999 090 505 341 730 816 × 2 = 1 + 0.999 999 999 998 181 010 683 461 632;
  • 20) 0.999 999 999 998 181 010 683 461 632 × 2 = 1 + 0.999 999 999 996 362 021 366 923 264;
  • 21) 0.999 999 999 996 362 021 366 923 264 × 2 = 1 + 0.999 999 999 992 724 042 733 846 528;
  • 22) 0.999 999 999 992 724 042 733 846 528 × 2 = 1 + 0.999 999 999 985 448 085 467 693 056;
  • 23) 0.999 999 999 985 448 085 467 693 056 × 2 = 1 + 0.999 999 999 970 896 170 935 386 112;
  • 24) 0.999 999 999 970 896 170 935 386 112 × 2 = 1 + 0.999 999 999 941 792 341 870 772 224;
  • 25) 0.999 999 999 941 792 341 870 772 224 × 2 = 1 + 0.999 999 999 883 584 683 741 544 448;
  • 26) 0.999 999 999 883 584 683 741 544 448 × 2 = 1 + 0.999 999 999 767 169 367 483 088 896;
  • 27) 0.999 999 999 767 169 367 483 088 896 × 2 = 1 + 0.999 999 999 534 338 734 966 177 792;
  • 28) 0.999 999 999 534 338 734 966 177 792 × 2 = 1 + 0.999 999 999 068 677 469 932 355 584;
  • 29) 0.999 999 999 068 677 469 932 355 584 × 2 = 1 + 0.999 999 998 137 354 939 864 711 168;
  • 30) 0.999 999 998 137 354 939 864 711 168 × 2 = 1 + 0.999 999 996 274 709 879 729 422 336;
  • 31) 0.999 999 996 274 709 879 729 422 336 × 2 = 1 + 0.999 999 992 549 419 759 458 844 672;
  • 32) 0.999 999 992 549 419 759 458 844 672 × 2 = 1 + 0.999 999 985 098 839 518 917 689 344;
  • 33) 0.999 999 985 098 839 518 917 689 344 × 2 = 1 + 0.999 999 970 197 679 037 835 378 688;
  • 34) 0.999 999 970 197 679 037 835 378 688 × 2 = 1 + 0.999 999 940 395 358 075 670 757 376;
  • 35) 0.999 999 940 395 358 075 670 757 376 × 2 = 1 + 0.999 999 880 790 716 151 341 514 752;
  • 36) 0.999 999 880 790 716 151 341 514 752 × 2 = 1 + 0.999 999 761 581 432 302 683 029 504;
  • 37) 0.999 999 761 581 432 302 683 029 504 × 2 = 1 + 0.999 999 523 162 864 605 366 059 008;
  • 38) 0.999 999 523 162 864 605 366 059 008 × 2 = 1 + 0.999 999 046 325 729 210 732 118 016;
  • 39) 0.999 999 046 325 729 210 732 118 016 × 2 = 1 + 0.999 998 092 651 458 421 464 236 032;
  • 40) 0.999 998 092 651 458 421 464 236 032 × 2 = 1 + 0.999 996 185 302 916 842 928 472 064;
  • 41) 0.999 996 185 302 916 842 928 472 064 × 2 = 1 + 0.999 992 370 605 833 685 856 944 128;
  • 42) 0.999 992 370 605 833 685 856 944 128 × 2 = 1 + 0.999 984 741 211 667 371 713 888 256;
  • 43) 0.999 984 741 211 667 371 713 888 256 × 2 = 1 + 0.999 969 482 423 334 743 427 776 512;
  • 44) 0.999 969 482 423 334 743 427 776 512 × 2 = 1 + 0.999 938 964 846 669 486 855 553 024;
  • 45) 0.999 938 964 846 669 486 855 553 024 × 2 = 1 + 0.999 877 929 693 338 973 711 106 048;
  • 46) 0.999 877 929 693 338 973 711 106 048 × 2 = 1 + 0.999 755 859 386 677 947 422 212 096;
  • 47) 0.999 755 859 386 677 947 422 212 096 × 2 = 1 + 0.999 511 718 773 355 894 844 424 192;
  • 48) 0.999 511 718 773 355 894 844 424 192 × 2 = 1 + 0.999 023 437 546 711 789 688 848 384;
  • 49) 0.999 023 437 546 711 789 688 848 384 × 2 = 1 + 0.998 046 875 093 423 579 377 696 768;
  • 50) 0.998 046 875 093 423 579 377 696 768 × 2 = 1 + 0.996 093 750 186 847 158 755 393 536;
  • 51) 0.996 093 750 186 847 158 755 393 536 × 2 = 1 + 0.992 187 500 373 694 317 510 787 072;
  • 52) 0.992 187 500 373 694 317 510 787 072 × 2 = 1 + 0.984 375 000 747 388 635 021 574 144;
  • 53) 0.984 375 000 747 388 635 021 574 144 × 2 = 1 + 0.968 750 001 494 777 270 043 148 288;
  • 54) 0.968 750 001 494 777 270 043 148 288 × 2 = 1 + 0.937 500 002 989 554 540 086 296 576;
  • 55) 0.937 500 002 989 554 540 086 296 576 × 2 = 1 + 0.875 000 005 979 109 080 172 593 152;
  • 56) 0.875 000 005 979 109 080 172 593 152 × 2 = 1 + 0.750 000 011 958 218 160 345 186 304;
  • 57) 0.750 000 011 958 218 160 345 186 304 × 2 = 1 + 0.500 000 023 916 436 320 690 372 608;
  • 58) 0.500 000 023 916 436 320 690 372 608 × 2 = 1 + 0.000 000 047 832 872 641 380 745 216;

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:


0.016 738 891 601 562 496 530 553 214(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 214(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:


0.016 738 891 601 562 496 530 553 214(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 -0.016 738 891 601 562 496 530 553 214 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 -0.016 738 891 601 562 496 530 553 173 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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