-0.016 738 891 601 562 496 530 553 151 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 553 151(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
-0.016 738 891 601 562 496 530 553 151(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:

|-0.016 738 891 601 562 496 530 553 151| = 0.016 738 891 601 562 496 530 553 151


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: 0.016 738 891 601 562 496 530 553 151.

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) 0.016 738 891 601 562 496 530 553 151 × 2 = 0 + 0.033 477 783 203 124 993 061 106 302;
  • 2) 0.033 477 783 203 124 993 061 106 302 × 2 = 0 + 0.066 955 566 406 249 986 122 212 604;
  • 3) 0.066 955 566 406 249 986 122 212 604 × 2 = 0 + 0.133 911 132 812 499 972 244 425 208;
  • 4) 0.133 911 132 812 499 972 244 425 208 × 2 = 0 + 0.267 822 265 624 999 944 488 850 416;
  • 5) 0.267 822 265 624 999 944 488 850 416 × 2 = 0 + 0.535 644 531 249 999 888 977 700 832;
  • 6) 0.535 644 531 249 999 888 977 700 832 × 2 = 1 + 0.071 289 062 499 999 777 955 401 664;
  • 7) 0.071 289 062 499 999 777 955 401 664 × 2 = 0 + 0.142 578 124 999 999 555 910 803 328;
  • 8) 0.142 578 124 999 999 555 910 803 328 × 2 = 0 + 0.285 156 249 999 999 111 821 606 656;
  • 9) 0.285 156 249 999 999 111 821 606 656 × 2 = 0 + 0.570 312 499 999 998 223 643 213 312;
  • 10) 0.570 312 499 999 998 223 643 213 312 × 2 = 1 + 0.140 624 999 999 996 447 286 426 624;
  • 11) 0.140 624 999 999 996 447 286 426 624 × 2 = 0 + 0.281 249 999 999 992 894 572 853 248;
  • 12) 0.281 249 999 999 992 894 572 853 248 × 2 = 0 + 0.562 499 999 999 985 789 145 706 496;
  • 13) 0.562 499 999 999 985 789 145 706 496 × 2 = 1 + 0.124 999 999 999 971 578 291 412 992;
  • 14) 0.124 999 999 999 971 578 291 412 992 × 2 = 0 + 0.249 999 999 999 943 156 582 825 984;
  • 15) 0.249 999 999 999 943 156 582 825 984 × 2 = 0 + 0.499 999 999 999 886 313 165 651 968;
  • 16) 0.499 999 999 999 886 313 165 651 968 × 2 = 0 + 0.999 999 999 999 772 626 331 303 936;
  • 17) 0.999 999 999 999 772 626 331 303 936 × 2 = 1 + 0.999 999 999 999 545 252 662 607 872;
  • 18) 0.999 999 999 999 545 252 662 607 872 × 2 = 1 + 0.999 999 999 999 090 505 325 215 744;
  • 19) 0.999 999 999 999 090 505 325 215 744 × 2 = 1 + 0.999 999 999 998 181 010 650 431 488;
  • 20) 0.999 999 999 998 181 010 650 431 488 × 2 = 1 + 0.999 999 999 996 362 021 300 862 976;
  • 21) 0.999 999 999 996 362 021 300 862 976 × 2 = 1 + 0.999 999 999 992 724 042 601 725 952;
  • 22) 0.999 999 999 992 724 042 601 725 952 × 2 = 1 + 0.999 999 999 985 448 085 203 451 904;
  • 23) 0.999 999 999 985 448 085 203 451 904 × 2 = 1 + 0.999 999 999 970 896 170 406 903 808;
  • 24) 0.999 999 999 970 896 170 406 903 808 × 2 = 1 + 0.999 999 999 941 792 340 813 807 616;
  • 25) 0.999 999 999 941 792 340 813 807 616 × 2 = 1 + 0.999 999 999 883 584 681 627 615 232;
  • 26) 0.999 999 999 883 584 681 627 615 232 × 2 = 1 + 0.999 999 999 767 169 363 255 230 464;
  • 27) 0.999 999 999 767 169 363 255 230 464 × 2 = 1 + 0.999 999 999 534 338 726 510 460 928;
  • 28) 0.999 999 999 534 338 726 510 460 928 × 2 = 1 + 0.999 999 999 068 677 453 020 921 856;
  • 29) 0.999 999 999 068 677 453 020 921 856 × 2 = 1 + 0.999 999 998 137 354 906 041 843 712;
  • 30) 0.999 999 998 137 354 906 041 843 712 × 2 = 1 + 0.999 999 996 274 709 812 083 687 424;
  • 31) 0.999 999 996 274 709 812 083 687 424 × 2 = 1 + 0.999 999 992 549 419 624 167 374 848;
  • 32) 0.999 999 992 549 419 624 167 374 848 × 2 = 1 + 0.999 999 985 098 839 248 334 749 696;
  • 33) 0.999 999 985 098 839 248 334 749 696 × 2 = 1 + 0.999 999 970 197 678 496 669 499 392;
  • 34) 0.999 999 970 197 678 496 669 499 392 × 2 = 1 + 0.999 999 940 395 356 993 338 998 784;
  • 35) 0.999 999 940 395 356 993 338 998 784 × 2 = 1 + 0.999 999 880 790 713 986 677 997 568;
  • 36) 0.999 999 880 790 713 986 677 997 568 × 2 = 1 + 0.999 999 761 581 427 973 355 995 136;
  • 37) 0.999 999 761 581 427 973 355 995 136 × 2 = 1 + 0.999 999 523 162 855 946 711 990 272;
  • 38) 0.999 999 523 162 855 946 711 990 272 × 2 = 1 + 0.999 999 046 325 711 893 423 980 544;
  • 39) 0.999 999 046 325 711 893 423 980 544 × 2 = 1 + 0.999 998 092 651 423 786 847 961 088;
  • 40) 0.999 998 092 651 423 786 847 961 088 × 2 = 1 + 0.999 996 185 302 847 573 695 922 176;
  • 41) 0.999 996 185 302 847 573 695 922 176 × 2 = 1 + 0.999 992 370 605 695 147 391 844 352;
  • 42) 0.999 992 370 605 695 147 391 844 352 × 2 = 1 + 0.999 984 741 211 390 294 783 688 704;
  • 43) 0.999 984 741 211 390 294 783 688 704 × 2 = 1 + 0.999 969 482 422 780 589 567 377 408;
  • 44) 0.999 969 482 422 780 589 567 377 408 × 2 = 1 + 0.999 938 964 845 561 179 134 754 816;
  • 45) 0.999 938 964 845 561 179 134 754 816 × 2 = 1 + 0.999 877 929 691 122 358 269 509 632;
  • 46) 0.999 877 929 691 122 358 269 509 632 × 2 = 1 + 0.999 755 859 382 244 716 539 019 264;
  • 47) 0.999 755 859 382 244 716 539 019 264 × 2 = 1 + 0.999 511 718 764 489 433 078 038 528;
  • 48) 0.999 511 718 764 489 433 078 038 528 × 2 = 1 + 0.999 023 437 528 978 866 156 077 056;
  • 49) 0.999 023 437 528 978 866 156 077 056 × 2 = 1 + 0.998 046 875 057 957 732 312 154 112;
  • 50) 0.998 046 875 057 957 732 312 154 112 × 2 = 1 + 0.996 093 750 115 915 464 624 308 224;
  • 51) 0.996 093 750 115 915 464 624 308 224 × 2 = 1 + 0.992 187 500 231 830 929 248 616 448;
  • 52) 0.992 187 500 231 830 929 248 616 448 × 2 = 1 + 0.984 375 000 463 661 858 497 232 896;
  • 53) 0.984 375 000 463 661 858 497 232 896 × 2 = 1 + 0.968 750 000 927 323 716 994 465 792;
  • 54) 0.968 750 000 927 323 716 994 465 792 × 2 = 1 + 0.937 500 001 854 647 433 988 931 584;
  • 55) 0.937 500 001 854 647 433 988 931 584 × 2 = 1 + 0.875 000 003 709 294 867 977 863 168;
  • 56) 0.875 000 003 709 294 867 977 863 168 × 2 = 1 + 0.750 000 007 418 589 735 955 726 336;
  • 57) 0.750 000 007 418 589 735 955 726 336 × 2 = 1 + 0.500 000 014 837 179 471 911 452 672;
  • 58) 0.500 000 014 837 179 471 911 452 672 × 2 = 1 + 0.000 000 029 674 358 943 822 905 344;

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:


0.016 738 891 601 562 496 530 553 151(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 553 151(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:


0.016 738 891 601 562 496 530 553 151(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 -0.016 738 891 601 562 496 530 553 151 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