-0.105 000 000 057 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.105 000 000 057(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
-0.105 000 000 057(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:

|-0.105 000 000 057| = 0.105 000 000 057


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: 0.105 000 000 057.

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) 0.105 000 000 057 × 2 = 0 + 0.210 000 000 114;
  • 2) 0.210 000 000 114 × 2 = 0 + 0.420 000 000 228;
  • 3) 0.420 000 000 228 × 2 = 0 + 0.840 000 000 456;
  • 4) 0.840 000 000 456 × 2 = 1 + 0.680 000 000 912;
  • 5) 0.680 000 000 912 × 2 = 1 + 0.360 000 001 824;
  • 6) 0.360 000 001 824 × 2 = 0 + 0.720 000 003 648;
  • 7) 0.720 000 003 648 × 2 = 1 + 0.440 000 007 296;
  • 8) 0.440 000 007 296 × 2 = 0 + 0.880 000 014 592;
  • 9) 0.880 000 014 592 × 2 = 1 + 0.760 000 029 184;
  • 10) 0.760 000 029 184 × 2 = 1 + 0.520 000 058 368;
  • 11) 0.520 000 058 368 × 2 = 1 + 0.040 000 116 736;
  • 12) 0.040 000 116 736 × 2 = 0 + 0.080 000 233 472;
  • 13) 0.080 000 233 472 × 2 = 0 + 0.160 000 466 944;
  • 14) 0.160 000 466 944 × 2 = 0 + 0.320 000 933 888;
  • 15) 0.320 000 933 888 × 2 = 0 + 0.640 001 867 776;
  • 16) 0.640 001 867 776 × 2 = 1 + 0.280 003 735 552;
  • 17) 0.280 003 735 552 × 2 = 0 + 0.560 007 471 104;
  • 18) 0.560 007 471 104 × 2 = 1 + 0.120 014 942 208;
  • 19) 0.120 014 942 208 × 2 = 0 + 0.240 029 884 416;
  • 20) 0.240 029 884 416 × 2 = 0 + 0.480 059 768 832;
  • 21) 0.480 059 768 832 × 2 = 0 + 0.960 119 537 664;
  • 22) 0.960 119 537 664 × 2 = 1 + 0.920 239 075 328;
  • 23) 0.920 239 075 328 × 2 = 1 + 0.840 478 150 656;
  • 24) 0.840 478 150 656 × 2 = 1 + 0.680 956 301 312;
  • 25) 0.680 956 301 312 × 2 = 1 + 0.361 912 602 624;
  • 26) 0.361 912 602 624 × 2 = 0 + 0.723 825 205 248;
  • 27) 0.723 825 205 248 × 2 = 1 + 0.447 650 410 496;
  • 28) 0.447 650 410 496 × 2 = 0 + 0.895 300 820 992;
  • 29) 0.895 300 820 992 × 2 = 1 + 0.790 601 641 984;
  • 30) 0.790 601 641 984 × 2 = 1 + 0.581 203 283 968;
  • 31) 0.581 203 283 968 × 2 = 1 + 0.162 406 567 936;
  • 32) 0.162 406 567 936 × 2 = 0 + 0.324 813 135 872;
  • 33) 0.324 813 135 872 × 2 = 0 + 0.649 626 271 744;
  • 34) 0.649 626 271 744 × 2 = 1 + 0.299 252 543 488;
  • 35) 0.299 252 543 488 × 2 = 0 + 0.598 505 086 976;
  • 36) 0.598 505 086 976 × 2 = 1 + 0.197 010 173 952;
  • 37) 0.197 010 173 952 × 2 = 0 + 0.394 020 347 904;
  • 38) 0.394 020 347 904 × 2 = 0 + 0.788 040 695 808;
  • 39) 0.788 040 695 808 × 2 = 1 + 0.576 081 391 616;
  • 40) 0.576 081 391 616 × 2 = 1 + 0.152 162 783 232;
  • 41) 0.152 162 783 232 × 2 = 0 + 0.304 325 566 464;
  • 42) 0.304 325 566 464 × 2 = 0 + 0.608 651 132 928;
  • 43) 0.608 651 132 928 × 2 = 1 + 0.217 302 265 856;
  • 44) 0.217 302 265 856 × 2 = 0 + 0.434 604 531 712;
  • 45) 0.434 604 531 712 × 2 = 0 + 0.869 209 063 424;
  • 46) 0.869 209 063 424 × 2 = 1 + 0.738 418 126 848;
  • 47) 0.738 418 126 848 × 2 = 1 + 0.476 836 253 696;
  • 48) 0.476 836 253 696 × 2 = 0 + 0.953 672 507 392;
  • 49) 0.953 672 507 392 × 2 = 1 + 0.907 345 014 784;
  • 50) 0.907 345 014 784 × 2 = 1 + 0.814 690 029 568;
  • 51) 0.814 690 029 568 × 2 = 1 + 0.629 380 059 136;
  • 52) 0.629 380 059 136 × 2 = 1 + 0.258 760 118 272;
  • 53) 0.258 760 118 272 × 2 = 0 + 0.517 520 236 544;
  • 54) 0.517 520 236 544 × 2 = 1 + 0.035 040 473 088;
  • 55) 0.035 040 473 088 × 2 = 0 + 0.070 080 946 176;
  • 56) 0.070 080 946 176 × 2 = 0 + 0.140 161 892 352;

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:


0.105 000 000 057(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100(2)

6. Positive number before normalization:

0.105 000 000 057(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.105 000 000 057(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100(2) =

0.0001 1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100(2) × 20 =

1.1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100(2) × 2-4


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):
1.1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 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) =

1019(10) =

011 1111 1011(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. 1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100 =

1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100


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 1011

Mantissa (52 bits) =
1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100


Decimal number -0.105 000 000 057 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 0111 1010 1110 0101 0011 0010 0110 1111 0100

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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