-0.016 738 891 601 562 496 530 606 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 606(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
-0.016 738 891 601 562 496 530 606(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:

|-0.016 738 891 601 562 496 530 606| = 0.016 738 891 601 562 496 530 606


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: 0.016 738 891 601 562 496 530 606.

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) 0.016 738 891 601 562 496 530 606 × 2 = 0 + 0.033 477 783 203 124 993 061 212;
  • 2) 0.033 477 783 203 124 993 061 212 × 2 = 0 + 0.066 955 566 406 249 986 122 424;
  • 3) 0.066 955 566 406 249 986 122 424 × 2 = 0 + 0.133 911 132 812 499 972 244 848;
  • 4) 0.133 911 132 812 499 972 244 848 × 2 = 0 + 0.267 822 265 624 999 944 489 696;
  • 5) 0.267 822 265 624 999 944 489 696 × 2 = 0 + 0.535 644 531 249 999 888 979 392;
  • 6) 0.535 644 531 249 999 888 979 392 × 2 = 1 + 0.071 289 062 499 999 777 958 784;
  • 7) 0.071 289 062 499 999 777 958 784 × 2 = 0 + 0.142 578 124 999 999 555 917 568;
  • 8) 0.142 578 124 999 999 555 917 568 × 2 = 0 + 0.285 156 249 999 999 111 835 136;
  • 9) 0.285 156 249 999 999 111 835 136 × 2 = 0 + 0.570 312 499 999 998 223 670 272;
  • 10) 0.570 312 499 999 998 223 670 272 × 2 = 1 + 0.140 624 999 999 996 447 340 544;
  • 11) 0.140 624 999 999 996 447 340 544 × 2 = 0 + 0.281 249 999 999 992 894 681 088;
  • 12) 0.281 249 999 999 992 894 681 088 × 2 = 0 + 0.562 499 999 999 985 789 362 176;
  • 13) 0.562 499 999 999 985 789 362 176 × 2 = 1 + 0.124 999 999 999 971 578 724 352;
  • 14) 0.124 999 999 999 971 578 724 352 × 2 = 0 + 0.249 999 999 999 943 157 448 704;
  • 15) 0.249 999 999 999 943 157 448 704 × 2 = 0 + 0.499 999 999 999 886 314 897 408;
  • 16) 0.499 999 999 999 886 314 897 408 × 2 = 0 + 0.999 999 999 999 772 629 794 816;
  • 17) 0.999 999 999 999 772 629 794 816 × 2 = 1 + 0.999 999 999 999 545 259 589 632;
  • 18) 0.999 999 999 999 545 259 589 632 × 2 = 1 + 0.999 999 999 999 090 519 179 264;
  • 19) 0.999 999 999 999 090 519 179 264 × 2 = 1 + 0.999 999 999 998 181 038 358 528;
  • 20) 0.999 999 999 998 181 038 358 528 × 2 = 1 + 0.999 999 999 996 362 076 717 056;
  • 21) 0.999 999 999 996 362 076 717 056 × 2 = 1 + 0.999 999 999 992 724 153 434 112;
  • 22) 0.999 999 999 992 724 153 434 112 × 2 = 1 + 0.999 999 999 985 448 306 868 224;
  • 23) 0.999 999 999 985 448 306 868 224 × 2 = 1 + 0.999 999 999 970 896 613 736 448;
  • 24) 0.999 999 999 970 896 613 736 448 × 2 = 1 + 0.999 999 999 941 793 227 472 896;
  • 25) 0.999 999 999 941 793 227 472 896 × 2 = 1 + 0.999 999 999 883 586 454 945 792;
  • 26) 0.999 999 999 883 586 454 945 792 × 2 = 1 + 0.999 999 999 767 172 909 891 584;
  • 27) 0.999 999 999 767 172 909 891 584 × 2 = 1 + 0.999 999 999 534 345 819 783 168;
  • 28) 0.999 999 999 534 345 819 783 168 × 2 = 1 + 0.999 999 999 068 691 639 566 336;
  • 29) 0.999 999 999 068 691 639 566 336 × 2 = 1 + 0.999 999 998 137 383 279 132 672;
  • 30) 0.999 999 998 137 383 279 132 672 × 2 = 1 + 0.999 999 996 274 766 558 265 344;
  • 31) 0.999 999 996 274 766 558 265 344 × 2 = 1 + 0.999 999 992 549 533 116 530 688;
  • 32) 0.999 999 992 549 533 116 530 688 × 2 = 1 + 0.999 999 985 099 066 233 061 376;
  • 33) 0.999 999 985 099 066 233 061 376 × 2 = 1 + 0.999 999 970 198 132 466 122 752;
  • 34) 0.999 999 970 198 132 466 122 752 × 2 = 1 + 0.999 999 940 396 264 932 245 504;
  • 35) 0.999 999 940 396 264 932 245 504 × 2 = 1 + 0.999 999 880 792 529 864 491 008;
  • 36) 0.999 999 880 792 529 864 491 008 × 2 = 1 + 0.999 999 761 585 059 728 982 016;
  • 37) 0.999 999 761 585 059 728 982 016 × 2 = 1 + 0.999 999 523 170 119 457 964 032;
  • 38) 0.999 999 523 170 119 457 964 032 × 2 = 1 + 0.999 999 046 340 238 915 928 064;
  • 39) 0.999 999 046 340 238 915 928 064 × 2 = 1 + 0.999 998 092 680 477 831 856 128;
  • 40) 0.999 998 092 680 477 831 856 128 × 2 = 1 + 0.999 996 185 360 955 663 712 256;
  • 41) 0.999 996 185 360 955 663 712 256 × 2 = 1 + 0.999 992 370 721 911 327 424 512;
  • 42) 0.999 992 370 721 911 327 424 512 × 2 = 1 + 0.999 984 741 443 822 654 849 024;
  • 43) 0.999 984 741 443 822 654 849 024 × 2 = 1 + 0.999 969 482 887 645 309 698 048;
  • 44) 0.999 969 482 887 645 309 698 048 × 2 = 1 + 0.999 938 965 775 290 619 396 096;
  • 45) 0.999 938 965 775 290 619 396 096 × 2 = 1 + 0.999 877 931 550 581 238 792 192;
  • 46) 0.999 877 931 550 581 238 792 192 × 2 = 1 + 0.999 755 863 101 162 477 584 384;
  • 47) 0.999 755 863 101 162 477 584 384 × 2 = 1 + 0.999 511 726 202 324 955 168 768;
  • 48) 0.999 511 726 202 324 955 168 768 × 2 = 1 + 0.999 023 452 404 649 910 337 536;
  • 49) 0.999 023 452 404 649 910 337 536 × 2 = 1 + 0.998 046 904 809 299 820 675 072;
  • 50) 0.998 046 904 809 299 820 675 072 × 2 = 1 + 0.996 093 809 618 599 641 350 144;
  • 51) 0.996 093 809 618 599 641 350 144 × 2 = 1 + 0.992 187 619 237 199 282 700 288;
  • 52) 0.992 187 619 237 199 282 700 288 × 2 = 1 + 0.984 375 238 474 398 565 400 576;
  • 53) 0.984 375 238 474 398 565 400 576 × 2 = 1 + 0.968 750 476 948 797 130 801 152;
  • 54) 0.968 750 476 948 797 130 801 152 × 2 = 1 + 0.937 500 953 897 594 261 602 304;
  • 55) 0.937 500 953 897 594 261 602 304 × 2 = 1 + 0.875 001 907 795 188 523 204 608;
  • 56) 0.875 001 907 795 188 523 204 608 × 2 = 1 + 0.750 003 815 590 377 046 409 216;
  • 57) 0.750 003 815 590 377 046 409 216 × 2 = 1 + 0.500 007 631 180 754 092 818 432;
  • 58) 0.500 007 631 180 754 092 818 432 × 2 = 1 + 0.000 015 262 361 508 185 636 864;

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:


0.016 738 891 601 562 496 530 606(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 606(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:


0.016 738 891 601 562 496 530 606(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 -0.016 738 891 601 562 496 530 606 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