-0.016 738 891 601 562 496 530 392 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 392(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
-0.016 738 891 601 562 496 530 392(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:

|-0.016 738 891 601 562 496 530 392| = 0.016 738 891 601 562 496 530 392


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: 0.016 738 891 601 562 496 530 392.

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) 0.016 738 891 601 562 496 530 392 × 2 = 0 + 0.033 477 783 203 124 993 060 784;
  • 2) 0.033 477 783 203 124 993 060 784 × 2 = 0 + 0.066 955 566 406 249 986 121 568;
  • 3) 0.066 955 566 406 249 986 121 568 × 2 = 0 + 0.133 911 132 812 499 972 243 136;
  • 4) 0.133 911 132 812 499 972 243 136 × 2 = 0 + 0.267 822 265 624 999 944 486 272;
  • 5) 0.267 822 265 624 999 944 486 272 × 2 = 0 + 0.535 644 531 249 999 888 972 544;
  • 6) 0.535 644 531 249 999 888 972 544 × 2 = 1 + 0.071 289 062 499 999 777 945 088;
  • 7) 0.071 289 062 499 999 777 945 088 × 2 = 0 + 0.142 578 124 999 999 555 890 176;
  • 8) 0.142 578 124 999 999 555 890 176 × 2 = 0 + 0.285 156 249 999 999 111 780 352;
  • 9) 0.285 156 249 999 999 111 780 352 × 2 = 0 + 0.570 312 499 999 998 223 560 704;
  • 10) 0.570 312 499 999 998 223 560 704 × 2 = 1 + 0.140 624 999 999 996 447 121 408;
  • 11) 0.140 624 999 999 996 447 121 408 × 2 = 0 + 0.281 249 999 999 992 894 242 816;
  • 12) 0.281 249 999 999 992 894 242 816 × 2 = 0 + 0.562 499 999 999 985 788 485 632;
  • 13) 0.562 499 999 999 985 788 485 632 × 2 = 1 + 0.124 999 999 999 971 576 971 264;
  • 14) 0.124 999 999 999 971 576 971 264 × 2 = 0 + 0.249 999 999 999 943 153 942 528;
  • 15) 0.249 999 999 999 943 153 942 528 × 2 = 0 + 0.499 999 999 999 886 307 885 056;
  • 16) 0.499 999 999 999 886 307 885 056 × 2 = 0 + 0.999 999 999 999 772 615 770 112;
  • 17) 0.999 999 999 999 772 615 770 112 × 2 = 1 + 0.999 999 999 999 545 231 540 224;
  • 18) 0.999 999 999 999 545 231 540 224 × 2 = 1 + 0.999 999 999 999 090 463 080 448;
  • 19) 0.999 999 999 999 090 463 080 448 × 2 = 1 + 0.999 999 999 998 180 926 160 896;
  • 20) 0.999 999 999 998 180 926 160 896 × 2 = 1 + 0.999 999 999 996 361 852 321 792;
  • 21) 0.999 999 999 996 361 852 321 792 × 2 = 1 + 0.999 999 999 992 723 704 643 584;
  • 22) 0.999 999 999 992 723 704 643 584 × 2 = 1 + 0.999 999 999 985 447 409 287 168;
  • 23) 0.999 999 999 985 447 409 287 168 × 2 = 1 + 0.999 999 999 970 894 818 574 336;
  • 24) 0.999 999 999 970 894 818 574 336 × 2 = 1 + 0.999 999 999 941 789 637 148 672;
  • 25) 0.999 999 999 941 789 637 148 672 × 2 = 1 + 0.999 999 999 883 579 274 297 344;
  • 26) 0.999 999 999 883 579 274 297 344 × 2 = 1 + 0.999 999 999 767 158 548 594 688;
  • 27) 0.999 999 999 767 158 548 594 688 × 2 = 1 + 0.999 999 999 534 317 097 189 376;
  • 28) 0.999 999 999 534 317 097 189 376 × 2 = 1 + 0.999 999 999 068 634 194 378 752;
  • 29) 0.999 999 999 068 634 194 378 752 × 2 = 1 + 0.999 999 998 137 268 388 757 504;
  • 30) 0.999 999 998 137 268 388 757 504 × 2 = 1 + 0.999 999 996 274 536 777 515 008;
  • 31) 0.999 999 996 274 536 777 515 008 × 2 = 1 + 0.999 999 992 549 073 555 030 016;
  • 32) 0.999 999 992 549 073 555 030 016 × 2 = 1 + 0.999 999 985 098 147 110 060 032;
  • 33) 0.999 999 985 098 147 110 060 032 × 2 = 1 + 0.999 999 970 196 294 220 120 064;
  • 34) 0.999 999 970 196 294 220 120 064 × 2 = 1 + 0.999 999 940 392 588 440 240 128;
  • 35) 0.999 999 940 392 588 440 240 128 × 2 = 1 + 0.999 999 880 785 176 880 480 256;
  • 36) 0.999 999 880 785 176 880 480 256 × 2 = 1 + 0.999 999 761 570 353 760 960 512;
  • 37) 0.999 999 761 570 353 760 960 512 × 2 = 1 + 0.999 999 523 140 707 521 921 024;
  • 38) 0.999 999 523 140 707 521 921 024 × 2 = 1 + 0.999 999 046 281 415 043 842 048;
  • 39) 0.999 999 046 281 415 043 842 048 × 2 = 1 + 0.999 998 092 562 830 087 684 096;
  • 40) 0.999 998 092 562 830 087 684 096 × 2 = 1 + 0.999 996 185 125 660 175 368 192;
  • 41) 0.999 996 185 125 660 175 368 192 × 2 = 1 + 0.999 992 370 251 320 350 736 384;
  • 42) 0.999 992 370 251 320 350 736 384 × 2 = 1 + 0.999 984 740 502 640 701 472 768;
  • 43) 0.999 984 740 502 640 701 472 768 × 2 = 1 + 0.999 969 481 005 281 402 945 536;
  • 44) 0.999 969 481 005 281 402 945 536 × 2 = 1 + 0.999 938 962 010 562 805 891 072;
  • 45) 0.999 938 962 010 562 805 891 072 × 2 = 1 + 0.999 877 924 021 125 611 782 144;
  • 46) 0.999 877 924 021 125 611 782 144 × 2 = 1 + 0.999 755 848 042 251 223 564 288;
  • 47) 0.999 755 848 042 251 223 564 288 × 2 = 1 + 0.999 511 696 084 502 447 128 576;
  • 48) 0.999 511 696 084 502 447 128 576 × 2 = 1 + 0.999 023 392 169 004 894 257 152;
  • 49) 0.999 023 392 169 004 894 257 152 × 2 = 1 + 0.998 046 784 338 009 788 514 304;
  • 50) 0.998 046 784 338 009 788 514 304 × 2 = 1 + 0.996 093 568 676 019 577 028 608;
  • 51) 0.996 093 568 676 019 577 028 608 × 2 = 1 + 0.992 187 137 352 039 154 057 216;
  • 52) 0.992 187 137 352 039 154 057 216 × 2 = 1 + 0.984 374 274 704 078 308 114 432;
  • 53) 0.984 374 274 704 078 308 114 432 × 2 = 1 + 0.968 748 549 408 156 616 228 864;
  • 54) 0.968 748 549 408 156 616 228 864 × 2 = 1 + 0.937 497 098 816 313 232 457 728;
  • 55) 0.937 497 098 816 313 232 457 728 × 2 = 1 + 0.874 994 197 632 626 464 915 456;
  • 56) 0.874 994 197 632 626 464 915 456 × 2 = 1 + 0.749 988 395 265 252 929 830 912;
  • 57) 0.749 988 395 265 252 929 830 912 × 2 = 1 + 0.499 976 790 530 505 859 661 824;
  • 58) 0.499 976 790 530 505 859 661 824 × 2 = 0 + 0.999 953 581 061 011 719 323 648;

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:


0.016 738 891 601 562 496 530 392(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 392(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:


0.016 738 891 601 562 496 530 392(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 -0.016 738 891 601 562 496 530 392 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