-0.016 738 891 601 562 496 530 902 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 902(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
-0.016 738 891 601 562 496 530 902(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:

|-0.016 738 891 601 562 496 530 902| = 0.016 738 891 601 562 496 530 902


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: 0.016 738 891 601 562 496 530 902.

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) 0.016 738 891 601 562 496 530 902 × 2 = 0 + 0.033 477 783 203 124 993 061 804;
  • 2) 0.033 477 783 203 124 993 061 804 × 2 = 0 + 0.066 955 566 406 249 986 123 608;
  • 3) 0.066 955 566 406 249 986 123 608 × 2 = 0 + 0.133 911 132 812 499 972 247 216;
  • 4) 0.133 911 132 812 499 972 247 216 × 2 = 0 + 0.267 822 265 624 999 944 494 432;
  • 5) 0.267 822 265 624 999 944 494 432 × 2 = 0 + 0.535 644 531 249 999 888 988 864;
  • 6) 0.535 644 531 249 999 888 988 864 × 2 = 1 + 0.071 289 062 499 999 777 977 728;
  • 7) 0.071 289 062 499 999 777 977 728 × 2 = 0 + 0.142 578 124 999 999 555 955 456;
  • 8) 0.142 578 124 999 999 555 955 456 × 2 = 0 + 0.285 156 249 999 999 111 910 912;
  • 9) 0.285 156 249 999 999 111 910 912 × 2 = 0 + 0.570 312 499 999 998 223 821 824;
  • 10) 0.570 312 499 999 998 223 821 824 × 2 = 1 + 0.140 624 999 999 996 447 643 648;
  • 11) 0.140 624 999 999 996 447 643 648 × 2 = 0 + 0.281 249 999 999 992 895 287 296;
  • 12) 0.281 249 999 999 992 895 287 296 × 2 = 0 + 0.562 499 999 999 985 790 574 592;
  • 13) 0.562 499 999 999 985 790 574 592 × 2 = 1 + 0.124 999 999 999 971 581 149 184;
  • 14) 0.124 999 999 999 971 581 149 184 × 2 = 0 + 0.249 999 999 999 943 162 298 368;
  • 15) 0.249 999 999 999 943 162 298 368 × 2 = 0 + 0.499 999 999 999 886 324 596 736;
  • 16) 0.499 999 999 999 886 324 596 736 × 2 = 0 + 0.999 999 999 999 772 649 193 472;
  • 17) 0.999 999 999 999 772 649 193 472 × 2 = 1 + 0.999 999 999 999 545 298 386 944;
  • 18) 0.999 999 999 999 545 298 386 944 × 2 = 1 + 0.999 999 999 999 090 596 773 888;
  • 19) 0.999 999 999 999 090 596 773 888 × 2 = 1 + 0.999 999 999 998 181 193 547 776;
  • 20) 0.999 999 999 998 181 193 547 776 × 2 = 1 + 0.999 999 999 996 362 387 095 552;
  • 21) 0.999 999 999 996 362 387 095 552 × 2 = 1 + 0.999 999 999 992 724 774 191 104;
  • 22) 0.999 999 999 992 724 774 191 104 × 2 = 1 + 0.999 999 999 985 449 548 382 208;
  • 23) 0.999 999 999 985 449 548 382 208 × 2 = 1 + 0.999 999 999 970 899 096 764 416;
  • 24) 0.999 999 999 970 899 096 764 416 × 2 = 1 + 0.999 999 999 941 798 193 528 832;
  • 25) 0.999 999 999 941 798 193 528 832 × 2 = 1 + 0.999 999 999 883 596 387 057 664;
  • 26) 0.999 999 999 883 596 387 057 664 × 2 = 1 + 0.999 999 999 767 192 774 115 328;
  • 27) 0.999 999 999 767 192 774 115 328 × 2 = 1 + 0.999 999 999 534 385 548 230 656;
  • 28) 0.999 999 999 534 385 548 230 656 × 2 = 1 + 0.999 999 999 068 771 096 461 312;
  • 29) 0.999 999 999 068 771 096 461 312 × 2 = 1 + 0.999 999 998 137 542 192 922 624;
  • 30) 0.999 999 998 137 542 192 922 624 × 2 = 1 + 0.999 999 996 275 084 385 845 248;
  • 31) 0.999 999 996 275 084 385 845 248 × 2 = 1 + 0.999 999 992 550 168 771 690 496;
  • 32) 0.999 999 992 550 168 771 690 496 × 2 = 1 + 0.999 999 985 100 337 543 380 992;
  • 33) 0.999 999 985 100 337 543 380 992 × 2 = 1 + 0.999 999 970 200 675 086 761 984;
  • 34) 0.999 999 970 200 675 086 761 984 × 2 = 1 + 0.999 999 940 401 350 173 523 968;
  • 35) 0.999 999 940 401 350 173 523 968 × 2 = 1 + 0.999 999 880 802 700 347 047 936;
  • 36) 0.999 999 880 802 700 347 047 936 × 2 = 1 + 0.999 999 761 605 400 694 095 872;
  • 37) 0.999 999 761 605 400 694 095 872 × 2 = 1 + 0.999 999 523 210 801 388 191 744;
  • 38) 0.999 999 523 210 801 388 191 744 × 2 = 1 + 0.999 999 046 421 602 776 383 488;
  • 39) 0.999 999 046 421 602 776 383 488 × 2 = 1 + 0.999 998 092 843 205 552 766 976;
  • 40) 0.999 998 092 843 205 552 766 976 × 2 = 1 + 0.999 996 185 686 411 105 533 952;
  • 41) 0.999 996 185 686 411 105 533 952 × 2 = 1 + 0.999 992 371 372 822 211 067 904;
  • 42) 0.999 992 371 372 822 211 067 904 × 2 = 1 + 0.999 984 742 745 644 422 135 808;
  • 43) 0.999 984 742 745 644 422 135 808 × 2 = 1 + 0.999 969 485 491 288 844 271 616;
  • 44) 0.999 969 485 491 288 844 271 616 × 2 = 1 + 0.999 938 970 982 577 688 543 232;
  • 45) 0.999 938 970 982 577 688 543 232 × 2 = 1 + 0.999 877 941 965 155 377 086 464;
  • 46) 0.999 877 941 965 155 377 086 464 × 2 = 1 + 0.999 755 883 930 310 754 172 928;
  • 47) 0.999 755 883 930 310 754 172 928 × 2 = 1 + 0.999 511 767 860 621 508 345 856;
  • 48) 0.999 511 767 860 621 508 345 856 × 2 = 1 + 0.999 023 535 721 243 016 691 712;
  • 49) 0.999 023 535 721 243 016 691 712 × 2 = 1 + 0.998 047 071 442 486 033 383 424;
  • 50) 0.998 047 071 442 486 033 383 424 × 2 = 1 + 0.996 094 142 884 972 066 766 848;
  • 51) 0.996 094 142 884 972 066 766 848 × 2 = 1 + 0.992 188 285 769 944 133 533 696;
  • 52) 0.992 188 285 769 944 133 533 696 × 2 = 1 + 0.984 376 571 539 888 267 067 392;
  • 53) 0.984 376 571 539 888 267 067 392 × 2 = 1 + 0.968 753 143 079 776 534 134 784;
  • 54) 0.968 753 143 079 776 534 134 784 × 2 = 1 + 0.937 506 286 159 553 068 269 568;
  • 55) 0.937 506 286 159 553 068 269 568 × 2 = 1 + 0.875 012 572 319 106 136 539 136;
  • 56) 0.875 012 572 319 106 136 539 136 × 2 = 1 + 0.750 025 144 638 212 273 078 272;
  • 57) 0.750 025 144 638 212 273 078 272 × 2 = 1 + 0.500 050 289 276 424 546 156 544;
  • 58) 0.500 050 289 276 424 546 156 544 × 2 = 1 + 0.000 100 578 552 849 092 313 088;

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:


0.016 738 891 601 562 496 530 902(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 902(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:


0.016 738 891 601 562 496 530 902(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 -0.016 738 891 601 562 496 530 902 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