-0.016 738 891 601 562 496 530 497 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 497(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
-0.016 738 891 601 562 496 530 497(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:

|-0.016 738 891 601 562 496 530 497| = 0.016 738 891 601 562 496 530 497


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: 0.016 738 891 601 562 496 530 497.

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) 0.016 738 891 601 562 496 530 497 × 2 = 0 + 0.033 477 783 203 124 993 060 994;
  • 2) 0.033 477 783 203 124 993 060 994 × 2 = 0 + 0.066 955 566 406 249 986 121 988;
  • 3) 0.066 955 566 406 249 986 121 988 × 2 = 0 + 0.133 911 132 812 499 972 243 976;
  • 4) 0.133 911 132 812 499 972 243 976 × 2 = 0 + 0.267 822 265 624 999 944 487 952;
  • 5) 0.267 822 265 624 999 944 487 952 × 2 = 0 + 0.535 644 531 249 999 888 975 904;
  • 6) 0.535 644 531 249 999 888 975 904 × 2 = 1 + 0.071 289 062 499 999 777 951 808;
  • 7) 0.071 289 062 499 999 777 951 808 × 2 = 0 + 0.142 578 124 999 999 555 903 616;
  • 8) 0.142 578 124 999 999 555 903 616 × 2 = 0 + 0.285 156 249 999 999 111 807 232;
  • 9) 0.285 156 249 999 999 111 807 232 × 2 = 0 + 0.570 312 499 999 998 223 614 464;
  • 10) 0.570 312 499 999 998 223 614 464 × 2 = 1 + 0.140 624 999 999 996 447 228 928;
  • 11) 0.140 624 999 999 996 447 228 928 × 2 = 0 + 0.281 249 999 999 992 894 457 856;
  • 12) 0.281 249 999 999 992 894 457 856 × 2 = 0 + 0.562 499 999 999 985 788 915 712;
  • 13) 0.562 499 999 999 985 788 915 712 × 2 = 1 + 0.124 999 999 999 971 577 831 424;
  • 14) 0.124 999 999 999 971 577 831 424 × 2 = 0 + 0.249 999 999 999 943 155 662 848;
  • 15) 0.249 999 999 999 943 155 662 848 × 2 = 0 + 0.499 999 999 999 886 311 325 696;
  • 16) 0.499 999 999 999 886 311 325 696 × 2 = 0 + 0.999 999 999 999 772 622 651 392;
  • 17) 0.999 999 999 999 772 622 651 392 × 2 = 1 + 0.999 999 999 999 545 245 302 784;
  • 18) 0.999 999 999 999 545 245 302 784 × 2 = 1 + 0.999 999 999 999 090 490 605 568;
  • 19) 0.999 999 999 999 090 490 605 568 × 2 = 1 + 0.999 999 999 998 180 981 211 136;
  • 20) 0.999 999 999 998 180 981 211 136 × 2 = 1 + 0.999 999 999 996 361 962 422 272;
  • 21) 0.999 999 999 996 361 962 422 272 × 2 = 1 + 0.999 999 999 992 723 924 844 544;
  • 22) 0.999 999 999 992 723 924 844 544 × 2 = 1 + 0.999 999 999 985 447 849 689 088;
  • 23) 0.999 999 999 985 447 849 689 088 × 2 = 1 + 0.999 999 999 970 895 699 378 176;
  • 24) 0.999 999 999 970 895 699 378 176 × 2 = 1 + 0.999 999 999 941 791 398 756 352;
  • 25) 0.999 999 999 941 791 398 756 352 × 2 = 1 + 0.999 999 999 883 582 797 512 704;
  • 26) 0.999 999 999 883 582 797 512 704 × 2 = 1 + 0.999 999 999 767 165 595 025 408;
  • 27) 0.999 999 999 767 165 595 025 408 × 2 = 1 + 0.999 999 999 534 331 190 050 816;
  • 28) 0.999 999 999 534 331 190 050 816 × 2 = 1 + 0.999 999 999 068 662 380 101 632;
  • 29) 0.999 999 999 068 662 380 101 632 × 2 = 1 + 0.999 999 998 137 324 760 203 264;
  • 30) 0.999 999 998 137 324 760 203 264 × 2 = 1 + 0.999 999 996 274 649 520 406 528;
  • 31) 0.999 999 996 274 649 520 406 528 × 2 = 1 + 0.999 999 992 549 299 040 813 056;
  • 32) 0.999 999 992 549 299 040 813 056 × 2 = 1 + 0.999 999 985 098 598 081 626 112;
  • 33) 0.999 999 985 098 598 081 626 112 × 2 = 1 + 0.999 999 970 197 196 163 252 224;
  • 34) 0.999 999 970 197 196 163 252 224 × 2 = 1 + 0.999 999 940 394 392 326 504 448;
  • 35) 0.999 999 940 394 392 326 504 448 × 2 = 1 + 0.999 999 880 788 784 653 008 896;
  • 36) 0.999 999 880 788 784 653 008 896 × 2 = 1 + 0.999 999 761 577 569 306 017 792;
  • 37) 0.999 999 761 577 569 306 017 792 × 2 = 1 + 0.999 999 523 155 138 612 035 584;
  • 38) 0.999 999 523 155 138 612 035 584 × 2 = 1 + 0.999 999 046 310 277 224 071 168;
  • 39) 0.999 999 046 310 277 224 071 168 × 2 = 1 + 0.999 998 092 620 554 448 142 336;
  • 40) 0.999 998 092 620 554 448 142 336 × 2 = 1 + 0.999 996 185 241 108 896 284 672;
  • 41) 0.999 996 185 241 108 896 284 672 × 2 = 1 + 0.999 992 370 482 217 792 569 344;
  • 42) 0.999 992 370 482 217 792 569 344 × 2 = 1 + 0.999 984 740 964 435 585 138 688;
  • 43) 0.999 984 740 964 435 585 138 688 × 2 = 1 + 0.999 969 481 928 871 170 277 376;
  • 44) 0.999 969 481 928 871 170 277 376 × 2 = 1 + 0.999 938 963 857 742 340 554 752;
  • 45) 0.999 938 963 857 742 340 554 752 × 2 = 1 + 0.999 877 927 715 484 681 109 504;
  • 46) 0.999 877 927 715 484 681 109 504 × 2 = 1 + 0.999 755 855 430 969 362 219 008;
  • 47) 0.999 755 855 430 969 362 219 008 × 2 = 1 + 0.999 511 710 861 938 724 438 016;
  • 48) 0.999 511 710 861 938 724 438 016 × 2 = 1 + 0.999 023 421 723 877 448 876 032;
  • 49) 0.999 023 421 723 877 448 876 032 × 2 = 1 + 0.998 046 843 447 754 897 752 064;
  • 50) 0.998 046 843 447 754 897 752 064 × 2 = 1 + 0.996 093 686 895 509 795 504 128;
  • 51) 0.996 093 686 895 509 795 504 128 × 2 = 1 + 0.992 187 373 791 019 591 008 256;
  • 52) 0.992 187 373 791 019 591 008 256 × 2 = 1 + 0.984 374 747 582 039 182 016 512;
  • 53) 0.984 374 747 582 039 182 016 512 × 2 = 1 + 0.968 749 495 164 078 364 033 024;
  • 54) 0.968 749 495 164 078 364 033 024 × 2 = 1 + 0.937 498 990 328 156 728 066 048;
  • 55) 0.937 498 990 328 156 728 066 048 × 2 = 1 + 0.874 997 980 656 313 456 132 096;
  • 56) 0.874 997 980 656 313 456 132 096 × 2 = 1 + 0.749 995 961 312 626 912 264 192;
  • 57) 0.749 995 961 312 626 912 264 192 × 2 = 1 + 0.499 991 922 625 253 824 528 384;
  • 58) 0.499 991 922 625 253 824 528 384 × 2 = 0 + 0.999 983 845 250 507 649 056 768;

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:


0.016 738 891 601 562 496 530 497(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 497(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:


0.016 738 891 601 562 496 530 497(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 -0.016 738 891 601 562 496 530 497 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 -0.016 738 891 601 562 496 530 482 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