-0.016 738 891 601 562 496 532 71 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 532 71(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
-0.016 738 891 601 562 496 532 71(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:

|-0.016 738 891 601 562 496 532 71| = 0.016 738 891 601 562 496 532 71


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: 0.016 738 891 601 562 496 532 71.

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) 0.016 738 891 601 562 496 532 71 × 2 = 0 + 0.033 477 783 203 124 993 065 42;
  • 2) 0.033 477 783 203 124 993 065 42 × 2 = 0 + 0.066 955 566 406 249 986 130 84;
  • 3) 0.066 955 566 406 249 986 130 84 × 2 = 0 + 0.133 911 132 812 499 972 261 68;
  • 4) 0.133 911 132 812 499 972 261 68 × 2 = 0 + 0.267 822 265 624 999 944 523 36;
  • 5) 0.267 822 265 624 999 944 523 36 × 2 = 0 + 0.535 644 531 249 999 889 046 72;
  • 6) 0.535 644 531 249 999 889 046 72 × 2 = 1 + 0.071 289 062 499 999 778 093 44;
  • 7) 0.071 289 062 499 999 778 093 44 × 2 = 0 + 0.142 578 124 999 999 556 186 88;
  • 8) 0.142 578 124 999 999 556 186 88 × 2 = 0 + 0.285 156 249 999 999 112 373 76;
  • 9) 0.285 156 249 999 999 112 373 76 × 2 = 0 + 0.570 312 499 999 998 224 747 52;
  • 10) 0.570 312 499 999 998 224 747 52 × 2 = 1 + 0.140 624 999 999 996 449 495 04;
  • 11) 0.140 624 999 999 996 449 495 04 × 2 = 0 + 0.281 249 999 999 992 898 990 08;
  • 12) 0.281 249 999 999 992 898 990 08 × 2 = 0 + 0.562 499 999 999 985 797 980 16;
  • 13) 0.562 499 999 999 985 797 980 16 × 2 = 1 + 0.124 999 999 999 971 595 960 32;
  • 14) 0.124 999 999 999 971 595 960 32 × 2 = 0 + 0.249 999 999 999 943 191 920 64;
  • 15) 0.249 999 999 999 943 191 920 64 × 2 = 0 + 0.499 999 999 999 886 383 841 28;
  • 16) 0.499 999 999 999 886 383 841 28 × 2 = 0 + 0.999 999 999 999 772 767 682 56;
  • 17) 0.999 999 999 999 772 767 682 56 × 2 = 1 + 0.999 999 999 999 545 535 365 12;
  • 18) 0.999 999 999 999 545 535 365 12 × 2 = 1 + 0.999 999 999 999 091 070 730 24;
  • 19) 0.999 999 999 999 091 070 730 24 × 2 = 1 + 0.999 999 999 998 182 141 460 48;
  • 20) 0.999 999 999 998 182 141 460 48 × 2 = 1 + 0.999 999 999 996 364 282 920 96;
  • 21) 0.999 999 999 996 364 282 920 96 × 2 = 1 + 0.999 999 999 992 728 565 841 92;
  • 22) 0.999 999 999 992 728 565 841 92 × 2 = 1 + 0.999 999 999 985 457 131 683 84;
  • 23) 0.999 999 999 985 457 131 683 84 × 2 = 1 + 0.999 999 999 970 914 263 367 68;
  • 24) 0.999 999 999 970 914 263 367 68 × 2 = 1 + 0.999 999 999 941 828 526 735 36;
  • 25) 0.999 999 999 941 828 526 735 36 × 2 = 1 + 0.999 999 999 883 657 053 470 72;
  • 26) 0.999 999 999 883 657 053 470 72 × 2 = 1 + 0.999 999 999 767 314 106 941 44;
  • 27) 0.999 999 999 767 314 106 941 44 × 2 = 1 + 0.999 999 999 534 628 213 882 88;
  • 28) 0.999 999 999 534 628 213 882 88 × 2 = 1 + 0.999 999 999 069 256 427 765 76;
  • 29) 0.999 999 999 069 256 427 765 76 × 2 = 1 + 0.999 999 998 138 512 855 531 52;
  • 30) 0.999 999 998 138 512 855 531 52 × 2 = 1 + 0.999 999 996 277 025 711 063 04;
  • 31) 0.999 999 996 277 025 711 063 04 × 2 = 1 + 0.999 999 992 554 051 422 126 08;
  • 32) 0.999 999 992 554 051 422 126 08 × 2 = 1 + 0.999 999 985 108 102 844 252 16;
  • 33) 0.999 999 985 108 102 844 252 16 × 2 = 1 + 0.999 999 970 216 205 688 504 32;
  • 34) 0.999 999 970 216 205 688 504 32 × 2 = 1 + 0.999 999 940 432 411 377 008 64;
  • 35) 0.999 999 940 432 411 377 008 64 × 2 = 1 + 0.999 999 880 864 822 754 017 28;
  • 36) 0.999 999 880 864 822 754 017 28 × 2 = 1 + 0.999 999 761 729 645 508 034 56;
  • 37) 0.999 999 761 729 645 508 034 56 × 2 = 1 + 0.999 999 523 459 291 016 069 12;
  • 38) 0.999 999 523 459 291 016 069 12 × 2 = 1 + 0.999 999 046 918 582 032 138 24;
  • 39) 0.999 999 046 918 582 032 138 24 × 2 = 1 + 0.999 998 093 837 164 064 276 48;
  • 40) 0.999 998 093 837 164 064 276 48 × 2 = 1 + 0.999 996 187 674 328 128 552 96;
  • 41) 0.999 996 187 674 328 128 552 96 × 2 = 1 + 0.999 992 375 348 656 257 105 92;
  • 42) 0.999 992 375 348 656 257 105 92 × 2 = 1 + 0.999 984 750 697 312 514 211 84;
  • 43) 0.999 984 750 697 312 514 211 84 × 2 = 1 + 0.999 969 501 394 625 028 423 68;
  • 44) 0.999 969 501 394 625 028 423 68 × 2 = 1 + 0.999 939 002 789 250 056 847 36;
  • 45) 0.999 939 002 789 250 056 847 36 × 2 = 1 + 0.999 878 005 578 500 113 694 72;
  • 46) 0.999 878 005 578 500 113 694 72 × 2 = 1 + 0.999 756 011 157 000 227 389 44;
  • 47) 0.999 756 011 157 000 227 389 44 × 2 = 1 + 0.999 512 022 314 000 454 778 88;
  • 48) 0.999 512 022 314 000 454 778 88 × 2 = 1 + 0.999 024 044 628 000 909 557 76;
  • 49) 0.999 024 044 628 000 909 557 76 × 2 = 1 + 0.998 048 089 256 001 819 115 52;
  • 50) 0.998 048 089 256 001 819 115 52 × 2 = 1 + 0.996 096 178 512 003 638 231 04;
  • 51) 0.996 096 178 512 003 638 231 04 × 2 = 1 + 0.992 192 357 024 007 276 462 08;
  • 52) 0.992 192 357 024 007 276 462 08 × 2 = 1 + 0.984 384 714 048 014 552 924 16;
  • 53) 0.984 384 714 048 014 552 924 16 × 2 = 1 + 0.968 769 428 096 029 105 848 32;
  • 54) 0.968 769 428 096 029 105 848 32 × 2 = 1 + 0.937 538 856 192 058 211 696 64;
  • 55) 0.937 538 856 192 058 211 696 64 × 2 = 1 + 0.875 077 712 384 116 423 393 28;
  • 56) 0.875 077 712 384 116 423 393 28 × 2 = 1 + 0.750 155 424 768 232 846 786 56;
  • 57) 0.750 155 424 768 232 846 786 56 × 2 = 1 + 0.500 310 849 536 465 693 573 12;
  • 58) 0.500 310 849 536 465 693 573 12 × 2 = 1 + 0.000 621 699 072 931 387 146 24;

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:


0.016 738 891 601 562 496 532 71(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 532 71(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:


0.016 738 891 601 562 496 532 71(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 -0.016 738 891 601 562 496 532 71 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