-0.016 738 891 601 562 496 530 553 102 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 102(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
-0.016 738 891 601 562 496 530 553 102(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:

|-0.016 738 891 601 562 496 530 553 102| = 0.016 738 891 601 562 496 530 553 102


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: 0.016 738 891 601 562 496 530 553 102.

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) 0.016 738 891 601 562 496 530 553 102 × 2 = 0 + 0.033 477 783 203 124 993 061 106 204;
  • 2) 0.033 477 783 203 124 993 061 106 204 × 2 = 0 + 0.066 955 566 406 249 986 122 212 408;
  • 3) 0.066 955 566 406 249 986 122 212 408 × 2 = 0 + 0.133 911 132 812 499 972 244 424 816;
  • 4) 0.133 911 132 812 499 972 244 424 816 × 2 = 0 + 0.267 822 265 624 999 944 488 849 632;
  • 5) 0.267 822 265 624 999 944 488 849 632 × 2 = 0 + 0.535 644 531 249 999 888 977 699 264;
  • 6) 0.535 644 531 249 999 888 977 699 264 × 2 = 1 + 0.071 289 062 499 999 777 955 398 528;
  • 7) 0.071 289 062 499 999 777 955 398 528 × 2 = 0 + 0.142 578 124 999 999 555 910 797 056;
  • 8) 0.142 578 124 999 999 555 910 797 056 × 2 = 0 + 0.285 156 249 999 999 111 821 594 112;
  • 9) 0.285 156 249 999 999 111 821 594 112 × 2 = 0 + 0.570 312 499 999 998 223 643 188 224;
  • 10) 0.570 312 499 999 998 223 643 188 224 × 2 = 1 + 0.140 624 999 999 996 447 286 376 448;
  • 11) 0.140 624 999 999 996 447 286 376 448 × 2 = 0 + 0.281 249 999 999 992 894 572 752 896;
  • 12) 0.281 249 999 999 992 894 572 752 896 × 2 = 0 + 0.562 499 999 999 985 789 145 505 792;
  • 13) 0.562 499 999 999 985 789 145 505 792 × 2 = 1 + 0.124 999 999 999 971 578 291 011 584;
  • 14) 0.124 999 999 999 971 578 291 011 584 × 2 = 0 + 0.249 999 999 999 943 156 582 023 168;
  • 15) 0.249 999 999 999 943 156 582 023 168 × 2 = 0 + 0.499 999 999 999 886 313 164 046 336;
  • 16) 0.499 999 999 999 886 313 164 046 336 × 2 = 0 + 0.999 999 999 999 772 626 328 092 672;
  • 17) 0.999 999 999 999 772 626 328 092 672 × 2 = 1 + 0.999 999 999 999 545 252 656 185 344;
  • 18) 0.999 999 999 999 545 252 656 185 344 × 2 = 1 + 0.999 999 999 999 090 505 312 370 688;
  • 19) 0.999 999 999 999 090 505 312 370 688 × 2 = 1 + 0.999 999 999 998 181 010 624 741 376;
  • 20) 0.999 999 999 998 181 010 624 741 376 × 2 = 1 + 0.999 999 999 996 362 021 249 482 752;
  • 21) 0.999 999 999 996 362 021 249 482 752 × 2 = 1 + 0.999 999 999 992 724 042 498 965 504;
  • 22) 0.999 999 999 992 724 042 498 965 504 × 2 = 1 + 0.999 999 999 985 448 084 997 931 008;
  • 23) 0.999 999 999 985 448 084 997 931 008 × 2 = 1 + 0.999 999 999 970 896 169 995 862 016;
  • 24) 0.999 999 999 970 896 169 995 862 016 × 2 = 1 + 0.999 999 999 941 792 339 991 724 032;
  • 25) 0.999 999 999 941 792 339 991 724 032 × 2 = 1 + 0.999 999 999 883 584 679 983 448 064;
  • 26) 0.999 999 999 883 584 679 983 448 064 × 2 = 1 + 0.999 999 999 767 169 359 966 896 128;
  • 27) 0.999 999 999 767 169 359 966 896 128 × 2 = 1 + 0.999 999 999 534 338 719 933 792 256;
  • 28) 0.999 999 999 534 338 719 933 792 256 × 2 = 1 + 0.999 999 999 068 677 439 867 584 512;
  • 29) 0.999 999 999 068 677 439 867 584 512 × 2 = 1 + 0.999 999 998 137 354 879 735 169 024;
  • 30) 0.999 999 998 137 354 879 735 169 024 × 2 = 1 + 0.999 999 996 274 709 759 470 338 048;
  • 31) 0.999 999 996 274 709 759 470 338 048 × 2 = 1 + 0.999 999 992 549 419 518 940 676 096;
  • 32) 0.999 999 992 549 419 518 940 676 096 × 2 = 1 + 0.999 999 985 098 839 037 881 352 192;
  • 33) 0.999 999 985 098 839 037 881 352 192 × 2 = 1 + 0.999 999 970 197 678 075 762 704 384;
  • 34) 0.999 999 970 197 678 075 762 704 384 × 2 = 1 + 0.999 999 940 395 356 151 525 408 768;
  • 35) 0.999 999 940 395 356 151 525 408 768 × 2 = 1 + 0.999 999 880 790 712 303 050 817 536;
  • 36) 0.999 999 880 790 712 303 050 817 536 × 2 = 1 + 0.999 999 761 581 424 606 101 635 072;
  • 37) 0.999 999 761 581 424 606 101 635 072 × 2 = 1 + 0.999 999 523 162 849 212 203 270 144;
  • 38) 0.999 999 523 162 849 212 203 270 144 × 2 = 1 + 0.999 999 046 325 698 424 406 540 288;
  • 39) 0.999 999 046 325 698 424 406 540 288 × 2 = 1 + 0.999 998 092 651 396 848 813 080 576;
  • 40) 0.999 998 092 651 396 848 813 080 576 × 2 = 1 + 0.999 996 185 302 793 697 626 161 152;
  • 41) 0.999 996 185 302 793 697 626 161 152 × 2 = 1 + 0.999 992 370 605 587 395 252 322 304;
  • 42) 0.999 992 370 605 587 395 252 322 304 × 2 = 1 + 0.999 984 741 211 174 790 504 644 608;
  • 43) 0.999 984 741 211 174 790 504 644 608 × 2 = 1 + 0.999 969 482 422 349 581 009 289 216;
  • 44) 0.999 969 482 422 349 581 009 289 216 × 2 = 1 + 0.999 938 964 844 699 162 018 578 432;
  • 45) 0.999 938 964 844 699 162 018 578 432 × 2 = 1 + 0.999 877 929 689 398 324 037 156 864;
  • 46) 0.999 877 929 689 398 324 037 156 864 × 2 = 1 + 0.999 755 859 378 796 648 074 313 728;
  • 47) 0.999 755 859 378 796 648 074 313 728 × 2 = 1 + 0.999 511 718 757 593 296 148 627 456;
  • 48) 0.999 511 718 757 593 296 148 627 456 × 2 = 1 + 0.999 023 437 515 186 592 297 254 912;
  • 49) 0.999 023 437 515 186 592 297 254 912 × 2 = 1 + 0.998 046 875 030 373 184 594 509 824;
  • 50) 0.998 046 875 030 373 184 594 509 824 × 2 = 1 + 0.996 093 750 060 746 369 189 019 648;
  • 51) 0.996 093 750 060 746 369 189 019 648 × 2 = 1 + 0.992 187 500 121 492 738 378 039 296;
  • 52) 0.992 187 500 121 492 738 378 039 296 × 2 = 1 + 0.984 375 000 242 985 476 756 078 592;
  • 53) 0.984 375 000 242 985 476 756 078 592 × 2 = 1 + 0.968 750 000 485 970 953 512 157 184;
  • 54) 0.968 750 000 485 970 953 512 157 184 × 2 = 1 + 0.937 500 000 971 941 907 024 314 368;
  • 55) 0.937 500 000 971 941 907 024 314 368 × 2 = 1 + 0.875 000 001 943 883 814 048 628 736;
  • 56) 0.875 000 001 943 883 814 048 628 736 × 2 = 1 + 0.750 000 003 887 767 628 097 257 472;
  • 57) 0.750 000 003 887 767 628 097 257 472 × 2 = 1 + 0.500 000 007 775 535 256 194 514 944;
  • 58) 0.500 000 007 775 535 256 194 514 944 × 2 = 1 + 0.000 000 015 551 070 512 389 029 888;

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:


0.016 738 891 601 562 496 530 553 102(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 102(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:


0.016 738 891 601 562 496 530 553 102(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 -0.016 738 891 601 562 496 530 553 102 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

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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