-0.016 738 891 601 562 496 530 89 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 89(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 89(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 89| = 0.016 738 891 601 562 496 530 89


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 89.

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 89 × 2 = 0 + 0.033 477 783 203 124 993 061 78;
  • 2) 0.033 477 783 203 124 993 061 78 × 2 = 0 + 0.066 955 566 406 249 986 123 56;
  • 3) 0.066 955 566 406 249 986 123 56 × 2 = 0 + 0.133 911 132 812 499 972 247 12;
  • 4) 0.133 911 132 812 499 972 247 12 × 2 = 0 + 0.267 822 265 624 999 944 494 24;
  • 5) 0.267 822 265 624 999 944 494 24 × 2 = 0 + 0.535 644 531 249 999 888 988 48;
  • 6) 0.535 644 531 249 999 888 988 48 × 2 = 1 + 0.071 289 062 499 999 777 976 96;
  • 7) 0.071 289 062 499 999 777 976 96 × 2 = 0 + 0.142 578 124 999 999 555 953 92;
  • 8) 0.142 578 124 999 999 555 953 92 × 2 = 0 + 0.285 156 249 999 999 111 907 84;
  • 9) 0.285 156 249 999 999 111 907 84 × 2 = 0 + 0.570 312 499 999 998 223 815 68;
  • 10) 0.570 312 499 999 998 223 815 68 × 2 = 1 + 0.140 624 999 999 996 447 631 36;
  • 11) 0.140 624 999 999 996 447 631 36 × 2 = 0 + 0.281 249 999 999 992 895 262 72;
  • 12) 0.281 249 999 999 992 895 262 72 × 2 = 0 + 0.562 499 999 999 985 790 525 44;
  • 13) 0.562 499 999 999 985 790 525 44 × 2 = 1 + 0.124 999 999 999 971 581 050 88;
  • 14) 0.124 999 999 999 971 581 050 88 × 2 = 0 + 0.249 999 999 999 943 162 101 76;
  • 15) 0.249 999 999 999 943 162 101 76 × 2 = 0 + 0.499 999 999 999 886 324 203 52;
  • 16) 0.499 999 999 999 886 324 203 52 × 2 = 0 + 0.999 999 999 999 772 648 407 04;
  • 17) 0.999 999 999 999 772 648 407 04 × 2 = 1 + 0.999 999 999 999 545 296 814 08;
  • 18) 0.999 999 999 999 545 296 814 08 × 2 = 1 + 0.999 999 999 999 090 593 628 16;
  • 19) 0.999 999 999 999 090 593 628 16 × 2 = 1 + 0.999 999 999 998 181 187 256 32;
  • 20) 0.999 999 999 998 181 187 256 32 × 2 = 1 + 0.999 999 999 996 362 374 512 64;
  • 21) 0.999 999 999 996 362 374 512 64 × 2 = 1 + 0.999 999 999 992 724 749 025 28;
  • 22) 0.999 999 999 992 724 749 025 28 × 2 = 1 + 0.999 999 999 985 449 498 050 56;
  • 23) 0.999 999 999 985 449 498 050 56 × 2 = 1 + 0.999 999 999 970 898 996 101 12;
  • 24) 0.999 999 999 970 898 996 101 12 × 2 = 1 + 0.999 999 999 941 797 992 202 24;
  • 25) 0.999 999 999 941 797 992 202 24 × 2 = 1 + 0.999 999 999 883 595 984 404 48;
  • 26) 0.999 999 999 883 595 984 404 48 × 2 = 1 + 0.999 999 999 767 191 968 808 96;
  • 27) 0.999 999 999 767 191 968 808 96 × 2 = 1 + 0.999 999 999 534 383 937 617 92;
  • 28) 0.999 999 999 534 383 937 617 92 × 2 = 1 + 0.999 999 999 068 767 875 235 84;
  • 29) 0.999 999 999 068 767 875 235 84 × 2 = 1 + 0.999 999 998 137 535 750 471 68;
  • 30) 0.999 999 998 137 535 750 471 68 × 2 = 1 + 0.999 999 996 275 071 500 943 36;
  • 31) 0.999 999 996 275 071 500 943 36 × 2 = 1 + 0.999 999 992 550 143 001 886 72;
  • 32) 0.999 999 992 550 143 001 886 72 × 2 = 1 + 0.999 999 985 100 286 003 773 44;
  • 33) 0.999 999 985 100 286 003 773 44 × 2 = 1 + 0.999 999 970 200 572 007 546 88;
  • 34) 0.999 999 970 200 572 007 546 88 × 2 = 1 + 0.999 999 940 401 144 015 093 76;
  • 35) 0.999 999 940 401 144 015 093 76 × 2 = 1 + 0.999 999 880 802 288 030 187 52;
  • 36) 0.999 999 880 802 288 030 187 52 × 2 = 1 + 0.999 999 761 604 576 060 375 04;
  • 37) 0.999 999 761 604 576 060 375 04 × 2 = 1 + 0.999 999 523 209 152 120 750 08;
  • 38) 0.999 999 523 209 152 120 750 08 × 2 = 1 + 0.999 999 046 418 304 241 500 16;
  • 39) 0.999 999 046 418 304 241 500 16 × 2 = 1 + 0.999 998 092 836 608 483 000 32;
  • 40) 0.999 998 092 836 608 483 000 32 × 2 = 1 + 0.999 996 185 673 216 966 000 64;
  • 41) 0.999 996 185 673 216 966 000 64 × 2 = 1 + 0.999 992 371 346 433 932 001 28;
  • 42) 0.999 992 371 346 433 932 001 28 × 2 = 1 + 0.999 984 742 692 867 864 002 56;
  • 43) 0.999 984 742 692 867 864 002 56 × 2 = 1 + 0.999 969 485 385 735 728 005 12;
  • 44) 0.999 969 485 385 735 728 005 12 × 2 = 1 + 0.999 938 970 771 471 456 010 24;
  • 45) 0.999 938 970 771 471 456 010 24 × 2 = 1 + 0.999 877 941 542 942 912 020 48;
  • 46) 0.999 877 941 542 942 912 020 48 × 2 = 1 + 0.999 755 883 085 885 824 040 96;
  • 47) 0.999 755 883 085 885 824 040 96 × 2 = 1 + 0.999 511 766 171 771 648 081 92;
  • 48) 0.999 511 766 171 771 648 081 92 × 2 = 1 + 0.999 023 532 343 543 296 163 84;
  • 49) 0.999 023 532 343 543 296 163 84 × 2 = 1 + 0.998 047 064 687 086 592 327 68;
  • 50) 0.998 047 064 687 086 592 327 68 × 2 = 1 + 0.996 094 129 374 173 184 655 36;
  • 51) 0.996 094 129 374 173 184 655 36 × 2 = 1 + 0.992 188 258 748 346 369 310 72;
  • 52) 0.992 188 258 748 346 369 310 72 × 2 = 1 + 0.984 376 517 496 692 738 621 44;
  • 53) 0.984 376 517 496 692 738 621 44 × 2 = 1 + 0.968 753 034 993 385 477 242 88;
  • 54) 0.968 753 034 993 385 477 242 88 × 2 = 1 + 0.937 506 069 986 770 954 485 76;
  • 55) 0.937 506 069 986 770 954 485 76 × 2 = 1 + 0.875 012 139 973 541 908 971 52;
  • 56) 0.875 012 139 973 541 908 971 52 × 2 = 1 + 0.750 024 279 947 083 817 943 04;
  • 57) 0.750 024 279 947 083 817 943 04 × 2 = 1 + 0.500 048 559 894 167 635 886 08;
  • 58) 0.500 048 559 894 167 635 886 08 × 2 = 1 + 0.000 097 119 788 335 271 772 16;

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 89(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 89(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 89(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 89 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