-0.016 738 891 601 562 496 598 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 598(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
-0.016 738 891 601 562 496 598(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:

|-0.016 738 891 601 562 496 598| = 0.016 738 891 601 562 496 598


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: 0.016 738 891 601 562 496 598.

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) 0.016 738 891 601 562 496 598 × 2 = 0 + 0.033 477 783 203 124 993 196;
  • 2) 0.033 477 783 203 124 993 196 × 2 = 0 + 0.066 955 566 406 249 986 392;
  • 3) 0.066 955 566 406 249 986 392 × 2 = 0 + 0.133 911 132 812 499 972 784;
  • 4) 0.133 911 132 812 499 972 784 × 2 = 0 + 0.267 822 265 624 999 945 568;
  • 5) 0.267 822 265 624 999 945 568 × 2 = 0 + 0.535 644 531 249 999 891 136;
  • 6) 0.535 644 531 249 999 891 136 × 2 = 1 + 0.071 289 062 499 999 782 272;
  • 7) 0.071 289 062 499 999 782 272 × 2 = 0 + 0.142 578 124 999 999 564 544;
  • 8) 0.142 578 124 999 999 564 544 × 2 = 0 + 0.285 156 249 999 999 129 088;
  • 9) 0.285 156 249 999 999 129 088 × 2 = 0 + 0.570 312 499 999 998 258 176;
  • 10) 0.570 312 499 999 998 258 176 × 2 = 1 + 0.140 624 999 999 996 516 352;
  • 11) 0.140 624 999 999 996 516 352 × 2 = 0 + 0.281 249 999 999 993 032 704;
  • 12) 0.281 249 999 999 993 032 704 × 2 = 0 + 0.562 499 999 999 986 065 408;
  • 13) 0.562 499 999 999 986 065 408 × 2 = 1 + 0.124 999 999 999 972 130 816;
  • 14) 0.124 999 999 999 972 130 816 × 2 = 0 + 0.249 999 999 999 944 261 632;
  • 15) 0.249 999 999 999 944 261 632 × 2 = 0 + 0.499 999 999 999 888 523 264;
  • 16) 0.499 999 999 999 888 523 264 × 2 = 0 + 0.999 999 999 999 777 046 528;
  • 17) 0.999 999 999 999 777 046 528 × 2 = 1 + 0.999 999 999 999 554 093 056;
  • 18) 0.999 999 999 999 554 093 056 × 2 = 1 + 0.999 999 999 999 108 186 112;
  • 19) 0.999 999 999 999 108 186 112 × 2 = 1 + 0.999 999 999 998 216 372 224;
  • 20) 0.999 999 999 998 216 372 224 × 2 = 1 + 0.999 999 999 996 432 744 448;
  • 21) 0.999 999 999 996 432 744 448 × 2 = 1 + 0.999 999 999 992 865 488 896;
  • 22) 0.999 999 999 992 865 488 896 × 2 = 1 + 0.999 999 999 985 730 977 792;
  • 23) 0.999 999 999 985 730 977 792 × 2 = 1 + 0.999 999 999 971 461 955 584;
  • 24) 0.999 999 999 971 461 955 584 × 2 = 1 + 0.999 999 999 942 923 911 168;
  • 25) 0.999 999 999 942 923 911 168 × 2 = 1 + 0.999 999 999 885 847 822 336;
  • 26) 0.999 999 999 885 847 822 336 × 2 = 1 + 0.999 999 999 771 695 644 672;
  • 27) 0.999 999 999 771 695 644 672 × 2 = 1 + 0.999 999 999 543 391 289 344;
  • 28) 0.999 999 999 543 391 289 344 × 2 = 1 + 0.999 999 999 086 782 578 688;
  • 29) 0.999 999 999 086 782 578 688 × 2 = 1 + 0.999 999 998 173 565 157 376;
  • 30) 0.999 999 998 173 565 157 376 × 2 = 1 + 0.999 999 996 347 130 314 752;
  • 31) 0.999 999 996 347 130 314 752 × 2 = 1 + 0.999 999 992 694 260 629 504;
  • 32) 0.999 999 992 694 260 629 504 × 2 = 1 + 0.999 999 985 388 521 259 008;
  • 33) 0.999 999 985 388 521 259 008 × 2 = 1 + 0.999 999 970 777 042 518 016;
  • 34) 0.999 999 970 777 042 518 016 × 2 = 1 + 0.999 999 941 554 085 036 032;
  • 35) 0.999 999 941 554 085 036 032 × 2 = 1 + 0.999 999 883 108 170 072 064;
  • 36) 0.999 999 883 108 170 072 064 × 2 = 1 + 0.999 999 766 216 340 144 128;
  • 37) 0.999 999 766 216 340 144 128 × 2 = 1 + 0.999 999 532 432 680 288 256;
  • 38) 0.999 999 532 432 680 288 256 × 2 = 1 + 0.999 999 064 865 360 576 512;
  • 39) 0.999 999 064 865 360 576 512 × 2 = 1 + 0.999 998 129 730 721 153 024;
  • 40) 0.999 998 129 730 721 153 024 × 2 = 1 + 0.999 996 259 461 442 306 048;
  • 41) 0.999 996 259 461 442 306 048 × 2 = 1 + 0.999 992 518 922 884 612 096;
  • 42) 0.999 992 518 922 884 612 096 × 2 = 1 + 0.999 985 037 845 769 224 192;
  • 43) 0.999 985 037 845 769 224 192 × 2 = 1 + 0.999 970 075 691 538 448 384;
  • 44) 0.999 970 075 691 538 448 384 × 2 = 1 + 0.999 940 151 383 076 896 768;
  • 45) 0.999 940 151 383 076 896 768 × 2 = 1 + 0.999 880 302 766 153 793 536;
  • 46) 0.999 880 302 766 153 793 536 × 2 = 1 + 0.999 760 605 532 307 587 072;
  • 47) 0.999 760 605 532 307 587 072 × 2 = 1 + 0.999 521 211 064 615 174 144;
  • 48) 0.999 521 211 064 615 174 144 × 2 = 1 + 0.999 042 422 129 230 348 288;
  • 49) 0.999 042 422 129 230 348 288 × 2 = 1 + 0.998 084 844 258 460 696 576;
  • 50) 0.998 084 844 258 460 696 576 × 2 = 1 + 0.996 169 688 516 921 393 152;
  • 51) 0.996 169 688 516 921 393 152 × 2 = 1 + 0.992 339 377 033 842 786 304;
  • 52) 0.992 339 377 033 842 786 304 × 2 = 1 + 0.984 678 754 067 685 572 608;
  • 53) 0.984 678 754 067 685 572 608 × 2 = 1 + 0.969 357 508 135 371 145 216;
  • 54) 0.969 357 508 135 371 145 216 × 2 = 1 + 0.938 715 016 270 742 290 432;
  • 55) 0.938 715 016 270 742 290 432 × 2 = 1 + 0.877 430 032 541 484 580 864;
  • 56) 0.877 430 032 541 484 580 864 × 2 = 1 + 0.754 860 065 082 969 161 728;
  • 57) 0.754 860 065 082 969 161 728 × 2 = 1 + 0.509 720 130 165 938 323 456;
  • 58) 0.509 720 130 165 938 323 456 × 2 = 1 + 0.019 440 260 331 876 646 912;

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:


0.016 738 891 601 562 496 598(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 598(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:


0.016 738 891 601 562 496 598(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 -0.016 738 891 601 562 496 598 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