-0.016 738 891 601 562 496 530 387 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 387(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
-0.016 738 891 601 562 496 530 387(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:

|-0.016 738 891 601 562 496 530 387| = 0.016 738 891 601 562 496 530 387


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: 0.016 738 891 601 562 496 530 387.

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) 0.016 738 891 601 562 496 530 387 × 2 = 0 + 0.033 477 783 203 124 993 060 774;
  • 2) 0.033 477 783 203 124 993 060 774 × 2 = 0 + 0.066 955 566 406 249 986 121 548;
  • 3) 0.066 955 566 406 249 986 121 548 × 2 = 0 + 0.133 911 132 812 499 972 243 096;
  • 4) 0.133 911 132 812 499 972 243 096 × 2 = 0 + 0.267 822 265 624 999 944 486 192;
  • 5) 0.267 822 265 624 999 944 486 192 × 2 = 0 + 0.535 644 531 249 999 888 972 384;
  • 6) 0.535 644 531 249 999 888 972 384 × 2 = 1 + 0.071 289 062 499 999 777 944 768;
  • 7) 0.071 289 062 499 999 777 944 768 × 2 = 0 + 0.142 578 124 999 999 555 889 536;
  • 8) 0.142 578 124 999 999 555 889 536 × 2 = 0 + 0.285 156 249 999 999 111 779 072;
  • 9) 0.285 156 249 999 999 111 779 072 × 2 = 0 + 0.570 312 499 999 998 223 558 144;
  • 10) 0.570 312 499 999 998 223 558 144 × 2 = 1 + 0.140 624 999 999 996 447 116 288;
  • 11) 0.140 624 999 999 996 447 116 288 × 2 = 0 + 0.281 249 999 999 992 894 232 576;
  • 12) 0.281 249 999 999 992 894 232 576 × 2 = 0 + 0.562 499 999 999 985 788 465 152;
  • 13) 0.562 499 999 999 985 788 465 152 × 2 = 1 + 0.124 999 999 999 971 576 930 304;
  • 14) 0.124 999 999 999 971 576 930 304 × 2 = 0 + 0.249 999 999 999 943 153 860 608;
  • 15) 0.249 999 999 999 943 153 860 608 × 2 = 0 + 0.499 999 999 999 886 307 721 216;
  • 16) 0.499 999 999 999 886 307 721 216 × 2 = 0 + 0.999 999 999 999 772 615 442 432;
  • 17) 0.999 999 999 999 772 615 442 432 × 2 = 1 + 0.999 999 999 999 545 230 884 864;
  • 18) 0.999 999 999 999 545 230 884 864 × 2 = 1 + 0.999 999 999 999 090 461 769 728;
  • 19) 0.999 999 999 999 090 461 769 728 × 2 = 1 + 0.999 999 999 998 180 923 539 456;
  • 20) 0.999 999 999 998 180 923 539 456 × 2 = 1 + 0.999 999 999 996 361 847 078 912;
  • 21) 0.999 999 999 996 361 847 078 912 × 2 = 1 + 0.999 999 999 992 723 694 157 824;
  • 22) 0.999 999 999 992 723 694 157 824 × 2 = 1 + 0.999 999 999 985 447 388 315 648;
  • 23) 0.999 999 999 985 447 388 315 648 × 2 = 1 + 0.999 999 999 970 894 776 631 296;
  • 24) 0.999 999 999 970 894 776 631 296 × 2 = 1 + 0.999 999 999 941 789 553 262 592;
  • 25) 0.999 999 999 941 789 553 262 592 × 2 = 1 + 0.999 999 999 883 579 106 525 184;
  • 26) 0.999 999 999 883 579 106 525 184 × 2 = 1 + 0.999 999 999 767 158 213 050 368;
  • 27) 0.999 999 999 767 158 213 050 368 × 2 = 1 + 0.999 999 999 534 316 426 100 736;
  • 28) 0.999 999 999 534 316 426 100 736 × 2 = 1 + 0.999 999 999 068 632 852 201 472;
  • 29) 0.999 999 999 068 632 852 201 472 × 2 = 1 + 0.999 999 998 137 265 704 402 944;
  • 30) 0.999 999 998 137 265 704 402 944 × 2 = 1 + 0.999 999 996 274 531 408 805 888;
  • 31) 0.999 999 996 274 531 408 805 888 × 2 = 1 + 0.999 999 992 549 062 817 611 776;
  • 32) 0.999 999 992 549 062 817 611 776 × 2 = 1 + 0.999 999 985 098 125 635 223 552;
  • 33) 0.999 999 985 098 125 635 223 552 × 2 = 1 + 0.999 999 970 196 251 270 447 104;
  • 34) 0.999 999 970 196 251 270 447 104 × 2 = 1 + 0.999 999 940 392 502 540 894 208;
  • 35) 0.999 999 940 392 502 540 894 208 × 2 = 1 + 0.999 999 880 785 005 081 788 416;
  • 36) 0.999 999 880 785 005 081 788 416 × 2 = 1 + 0.999 999 761 570 010 163 576 832;
  • 37) 0.999 999 761 570 010 163 576 832 × 2 = 1 + 0.999 999 523 140 020 327 153 664;
  • 38) 0.999 999 523 140 020 327 153 664 × 2 = 1 + 0.999 999 046 280 040 654 307 328;
  • 39) 0.999 999 046 280 040 654 307 328 × 2 = 1 + 0.999 998 092 560 081 308 614 656;
  • 40) 0.999 998 092 560 081 308 614 656 × 2 = 1 + 0.999 996 185 120 162 617 229 312;
  • 41) 0.999 996 185 120 162 617 229 312 × 2 = 1 + 0.999 992 370 240 325 234 458 624;
  • 42) 0.999 992 370 240 325 234 458 624 × 2 = 1 + 0.999 984 740 480 650 468 917 248;
  • 43) 0.999 984 740 480 650 468 917 248 × 2 = 1 + 0.999 969 480 961 300 937 834 496;
  • 44) 0.999 969 480 961 300 937 834 496 × 2 = 1 + 0.999 938 961 922 601 875 668 992;
  • 45) 0.999 938 961 922 601 875 668 992 × 2 = 1 + 0.999 877 923 845 203 751 337 984;
  • 46) 0.999 877 923 845 203 751 337 984 × 2 = 1 + 0.999 755 847 690 407 502 675 968;
  • 47) 0.999 755 847 690 407 502 675 968 × 2 = 1 + 0.999 511 695 380 815 005 351 936;
  • 48) 0.999 511 695 380 815 005 351 936 × 2 = 1 + 0.999 023 390 761 630 010 703 872;
  • 49) 0.999 023 390 761 630 010 703 872 × 2 = 1 + 0.998 046 781 523 260 021 407 744;
  • 50) 0.998 046 781 523 260 021 407 744 × 2 = 1 + 0.996 093 563 046 520 042 815 488;
  • 51) 0.996 093 563 046 520 042 815 488 × 2 = 1 + 0.992 187 126 093 040 085 630 976;
  • 52) 0.992 187 126 093 040 085 630 976 × 2 = 1 + 0.984 374 252 186 080 171 261 952;
  • 53) 0.984 374 252 186 080 171 261 952 × 2 = 1 + 0.968 748 504 372 160 342 523 904;
  • 54) 0.968 748 504 372 160 342 523 904 × 2 = 1 + 0.937 497 008 744 320 685 047 808;
  • 55) 0.937 497 008 744 320 685 047 808 × 2 = 1 + 0.874 994 017 488 641 370 095 616;
  • 56) 0.874 994 017 488 641 370 095 616 × 2 = 1 + 0.749 988 034 977 282 740 191 232;
  • 57) 0.749 988 034 977 282 740 191 232 × 2 = 1 + 0.499 976 069 954 565 480 382 464;
  • 58) 0.499 976 069 954 565 480 382 464 × 2 = 0 + 0.999 952 139 909 130 960 764 928;

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:


0.016 738 891 601 562 496 530 387(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 387(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:


0.016 738 891 601 562 496 530 387(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 -0.016 738 891 601 562 496 530 387 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