-0.016 738 891 601 562 496 530 488 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 488(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
-0.016 738 891 601 562 496 530 488(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:

|-0.016 738 891 601 562 496 530 488| = 0.016 738 891 601 562 496 530 488


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: 0.016 738 891 601 562 496 530 488.

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) 0.016 738 891 601 562 496 530 488 × 2 = 0 + 0.033 477 783 203 124 993 060 976;
  • 2) 0.033 477 783 203 124 993 060 976 × 2 = 0 + 0.066 955 566 406 249 986 121 952;
  • 3) 0.066 955 566 406 249 986 121 952 × 2 = 0 + 0.133 911 132 812 499 972 243 904;
  • 4) 0.133 911 132 812 499 972 243 904 × 2 = 0 + 0.267 822 265 624 999 944 487 808;
  • 5) 0.267 822 265 624 999 944 487 808 × 2 = 0 + 0.535 644 531 249 999 888 975 616;
  • 6) 0.535 644 531 249 999 888 975 616 × 2 = 1 + 0.071 289 062 499 999 777 951 232;
  • 7) 0.071 289 062 499 999 777 951 232 × 2 = 0 + 0.142 578 124 999 999 555 902 464;
  • 8) 0.142 578 124 999 999 555 902 464 × 2 = 0 + 0.285 156 249 999 999 111 804 928;
  • 9) 0.285 156 249 999 999 111 804 928 × 2 = 0 + 0.570 312 499 999 998 223 609 856;
  • 10) 0.570 312 499 999 998 223 609 856 × 2 = 1 + 0.140 624 999 999 996 447 219 712;
  • 11) 0.140 624 999 999 996 447 219 712 × 2 = 0 + 0.281 249 999 999 992 894 439 424;
  • 12) 0.281 249 999 999 992 894 439 424 × 2 = 0 + 0.562 499 999 999 985 788 878 848;
  • 13) 0.562 499 999 999 985 788 878 848 × 2 = 1 + 0.124 999 999 999 971 577 757 696;
  • 14) 0.124 999 999 999 971 577 757 696 × 2 = 0 + 0.249 999 999 999 943 155 515 392;
  • 15) 0.249 999 999 999 943 155 515 392 × 2 = 0 + 0.499 999 999 999 886 311 030 784;
  • 16) 0.499 999 999 999 886 311 030 784 × 2 = 0 + 0.999 999 999 999 772 622 061 568;
  • 17) 0.999 999 999 999 772 622 061 568 × 2 = 1 + 0.999 999 999 999 545 244 123 136;
  • 18) 0.999 999 999 999 545 244 123 136 × 2 = 1 + 0.999 999 999 999 090 488 246 272;
  • 19) 0.999 999 999 999 090 488 246 272 × 2 = 1 + 0.999 999 999 998 180 976 492 544;
  • 20) 0.999 999 999 998 180 976 492 544 × 2 = 1 + 0.999 999 999 996 361 952 985 088;
  • 21) 0.999 999 999 996 361 952 985 088 × 2 = 1 + 0.999 999 999 992 723 905 970 176;
  • 22) 0.999 999 999 992 723 905 970 176 × 2 = 1 + 0.999 999 999 985 447 811 940 352;
  • 23) 0.999 999 999 985 447 811 940 352 × 2 = 1 + 0.999 999 999 970 895 623 880 704;
  • 24) 0.999 999 999 970 895 623 880 704 × 2 = 1 + 0.999 999 999 941 791 247 761 408;
  • 25) 0.999 999 999 941 791 247 761 408 × 2 = 1 + 0.999 999 999 883 582 495 522 816;
  • 26) 0.999 999 999 883 582 495 522 816 × 2 = 1 + 0.999 999 999 767 164 991 045 632;
  • 27) 0.999 999 999 767 164 991 045 632 × 2 = 1 + 0.999 999 999 534 329 982 091 264;
  • 28) 0.999 999 999 534 329 982 091 264 × 2 = 1 + 0.999 999 999 068 659 964 182 528;
  • 29) 0.999 999 999 068 659 964 182 528 × 2 = 1 + 0.999 999 998 137 319 928 365 056;
  • 30) 0.999 999 998 137 319 928 365 056 × 2 = 1 + 0.999 999 996 274 639 856 730 112;
  • 31) 0.999 999 996 274 639 856 730 112 × 2 = 1 + 0.999 999 992 549 279 713 460 224;
  • 32) 0.999 999 992 549 279 713 460 224 × 2 = 1 + 0.999 999 985 098 559 426 920 448;
  • 33) 0.999 999 985 098 559 426 920 448 × 2 = 1 + 0.999 999 970 197 118 853 840 896;
  • 34) 0.999 999 970 197 118 853 840 896 × 2 = 1 + 0.999 999 940 394 237 707 681 792;
  • 35) 0.999 999 940 394 237 707 681 792 × 2 = 1 + 0.999 999 880 788 475 415 363 584;
  • 36) 0.999 999 880 788 475 415 363 584 × 2 = 1 + 0.999 999 761 576 950 830 727 168;
  • 37) 0.999 999 761 576 950 830 727 168 × 2 = 1 + 0.999 999 523 153 901 661 454 336;
  • 38) 0.999 999 523 153 901 661 454 336 × 2 = 1 + 0.999 999 046 307 803 322 908 672;
  • 39) 0.999 999 046 307 803 322 908 672 × 2 = 1 + 0.999 998 092 615 606 645 817 344;
  • 40) 0.999 998 092 615 606 645 817 344 × 2 = 1 + 0.999 996 185 231 213 291 634 688;
  • 41) 0.999 996 185 231 213 291 634 688 × 2 = 1 + 0.999 992 370 462 426 583 269 376;
  • 42) 0.999 992 370 462 426 583 269 376 × 2 = 1 + 0.999 984 740 924 853 166 538 752;
  • 43) 0.999 984 740 924 853 166 538 752 × 2 = 1 + 0.999 969 481 849 706 333 077 504;
  • 44) 0.999 969 481 849 706 333 077 504 × 2 = 1 + 0.999 938 963 699 412 666 155 008;
  • 45) 0.999 938 963 699 412 666 155 008 × 2 = 1 + 0.999 877 927 398 825 332 310 016;
  • 46) 0.999 877 927 398 825 332 310 016 × 2 = 1 + 0.999 755 854 797 650 664 620 032;
  • 47) 0.999 755 854 797 650 664 620 032 × 2 = 1 + 0.999 511 709 595 301 329 240 064;
  • 48) 0.999 511 709 595 301 329 240 064 × 2 = 1 + 0.999 023 419 190 602 658 480 128;
  • 49) 0.999 023 419 190 602 658 480 128 × 2 = 1 + 0.998 046 838 381 205 316 960 256;
  • 50) 0.998 046 838 381 205 316 960 256 × 2 = 1 + 0.996 093 676 762 410 633 920 512;
  • 51) 0.996 093 676 762 410 633 920 512 × 2 = 1 + 0.992 187 353 524 821 267 841 024;
  • 52) 0.992 187 353 524 821 267 841 024 × 2 = 1 + 0.984 374 707 049 642 535 682 048;
  • 53) 0.984 374 707 049 642 535 682 048 × 2 = 1 + 0.968 749 414 099 285 071 364 096;
  • 54) 0.968 749 414 099 285 071 364 096 × 2 = 1 + 0.937 498 828 198 570 142 728 192;
  • 55) 0.937 498 828 198 570 142 728 192 × 2 = 1 + 0.874 997 656 397 140 285 456 384;
  • 56) 0.874 997 656 397 140 285 456 384 × 2 = 1 + 0.749 995 312 794 280 570 912 768;
  • 57) 0.749 995 312 794 280 570 912 768 × 2 = 1 + 0.499 990 625 588 561 141 825 536;
  • 58) 0.499 990 625 588 561 141 825 536 × 2 = 0 + 0.999 981 251 177 122 283 651 072;

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:


0.016 738 891 601 562 496 530 488(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 488(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:


0.016 738 891 601 562 496 530 488(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 -0.016 738 891 601 562 496 530 488 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

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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