-0.000 000 000 008 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 008 5(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
-0.000 000 000 008 5(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:

|-0.000 000 000 008 5| = 0.000 000 000 008 5


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: 0.000 000 000 008 5.

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) 0.000 000 000 008 5 × 2 = 0 + 0.000 000 000 017;
  • 2) 0.000 000 000 017 × 2 = 0 + 0.000 000 000 034;
  • 3) 0.000 000 000 034 × 2 = 0 + 0.000 000 000 068;
  • 4) 0.000 000 000 068 × 2 = 0 + 0.000 000 000 136;
  • 5) 0.000 000 000 136 × 2 = 0 + 0.000 000 000 272;
  • 6) 0.000 000 000 272 × 2 = 0 + 0.000 000 000 544;
  • 7) 0.000 000 000 544 × 2 = 0 + 0.000 000 001 088;
  • 8) 0.000 000 001 088 × 2 = 0 + 0.000 000 002 176;
  • 9) 0.000 000 002 176 × 2 = 0 + 0.000 000 004 352;
  • 10) 0.000 000 004 352 × 2 = 0 + 0.000 000 008 704;
  • 11) 0.000 000 008 704 × 2 = 0 + 0.000 000 017 408;
  • 12) 0.000 000 017 408 × 2 = 0 + 0.000 000 034 816;
  • 13) 0.000 000 034 816 × 2 = 0 + 0.000 000 069 632;
  • 14) 0.000 000 069 632 × 2 = 0 + 0.000 000 139 264;
  • 15) 0.000 000 139 264 × 2 = 0 + 0.000 000 278 528;
  • 16) 0.000 000 278 528 × 2 = 0 + 0.000 000 557 056;
  • 17) 0.000 000 557 056 × 2 = 0 + 0.000 001 114 112;
  • 18) 0.000 001 114 112 × 2 = 0 + 0.000 002 228 224;
  • 19) 0.000 002 228 224 × 2 = 0 + 0.000 004 456 448;
  • 20) 0.000 004 456 448 × 2 = 0 + 0.000 008 912 896;
  • 21) 0.000 008 912 896 × 2 = 0 + 0.000 017 825 792;
  • 22) 0.000 017 825 792 × 2 = 0 + 0.000 035 651 584;
  • 23) 0.000 035 651 584 × 2 = 0 + 0.000 071 303 168;
  • 24) 0.000 071 303 168 × 2 = 0 + 0.000 142 606 336;
  • 25) 0.000 142 606 336 × 2 = 0 + 0.000 285 212 672;
  • 26) 0.000 285 212 672 × 2 = 0 + 0.000 570 425 344;
  • 27) 0.000 570 425 344 × 2 = 0 + 0.001 140 850 688;
  • 28) 0.001 140 850 688 × 2 = 0 + 0.002 281 701 376;
  • 29) 0.002 281 701 376 × 2 = 0 + 0.004 563 402 752;
  • 30) 0.004 563 402 752 × 2 = 0 + 0.009 126 805 504;
  • 31) 0.009 126 805 504 × 2 = 0 + 0.018 253 611 008;
  • 32) 0.018 253 611 008 × 2 = 0 + 0.036 507 222 016;
  • 33) 0.036 507 222 016 × 2 = 0 + 0.073 014 444 032;
  • 34) 0.073 014 444 032 × 2 = 0 + 0.146 028 888 064;
  • 35) 0.146 028 888 064 × 2 = 0 + 0.292 057 776 128;
  • 36) 0.292 057 776 128 × 2 = 0 + 0.584 115 552 256;
  • 37) 0.584 115 552 256 × 2 = 1 + 0.168 231 104 512;
  • 38) 0.168 231 104 512 × 2 = 0 + 0.336 462 209 024;
  • 39) 0.336 462 209 024 × 2 = 0 + 0.672 924 418 048;
  • 40) 0.672 924 418 048 × 2 = 1 + 0.345 848 836 096;
  • 41) 0.345 848 836 096 × 2 = 0 + 0.691 697 672 192;
  • 42) 0.691 697 672 192 × 2 = 1 + 0.383 395 344 384;
  • 43) 0.383 395 344 384 × 2 = 0 + 0.766 790 688 768;
  • 44) 0.766 790 688 768 × 2 = 1 + 0.533 581 377 536;
  • 45) 0.533 581 377 536 × 2 = 1 + 0.067 162 755 072;
  • 46) 0.067 162 755 072 × 2 = 0 + 0.134 325 510 144;
  • 47) 0.134 325 510 144 × 2 = 0 + 0.268 651 020 288;
  • 48) 0.268 651 020 288 × 2 = 0 + 0.537 302 040 576;
  • 49) 0.537 302 040 576 × 2 = 1 + 0.074 604 081 152;
  • 50) 0.074 604 081 152 × 2 = 0 + 0.149 208 162 304;
  • 51) 0.149 208 162 304 × 2 = 0 + 0.298 416 324 608;
  • 52) 0.298 416 324 608 × 2 = 0 + 0.596 832 649 216;
  • 53) 0.596 832 649 216 × 2 = 1 + 0.193 665 298 432;
  • 54) 0.193 665 298 432 × 2 = 0 + 0.387 330 596 864;
  • 55) 0.387 330 596 864 × 2 = 0 + 0.774 661 193 728;
  • 56) 0.774 661 193 728 × 2 = 1 + 0.549 322 387 456;
  • 57) 0.549 322 387 456 × 2 = 1 + 0.098 644 774 912;
  • 58) 0.098 644 774 912 × 2 = 0 + 0.197 289 549 824;
  • 59) 0.197 289 549 824 × 2 = 0 + 0.394 579 099 648;
  • 60) 0.394 579 099 648 × 2 = 0 + 0.789 158 199 296;
  • 61) 0.789 158 199 296 × 2 = 1 + 0.578 316 398 592;
  • 62) 0.578 316 398 592 × 2 = 1 + 0.156 632 797 184;
  • 63) 0.156 632 797 184 × 2 = 0 + 0.313 265 594 368;
  • 64) 0.313 265 594 368 × 2 = 0 + 0.626 531 188 736;
  • 65) 0.626 531 188 736 × 2 = 1 + 0.253 062 377 472;
  • 66) 0.253 062 377 472 × 2 = 0 + 0.506 124 754 944;
  • 67) 0.506 124 754 944 × 2 = 1 + 0.012 249 509 888;
  • 68) 0.012 249 509 888 × 2 = 0 + 0.024 499 019 776;
  • 69) 0.024 499 019 776 × 2 = 0 + 0.048 998 039 552;
  • 70) 0.048 998 039 552 × 2 = 0 + 0.097 996 079 104;
  • 71) 0.097 996 079 104 × 2 = 0 + 0.195 992 158 208;
  • 72) 0.195 992 158 208 × 2 = 0 + 0.391 984 316 416;
  • 73) 0.391 984 316 416 × 2 = 0 + 0.783 968 632 832;
  • 74) 0.783 968 632 832 × 2 = 1 + 0.567 937 265 664;
  • 75) 0.567 937 265 664 × 2 = 1 + 0.135 874 531 328;
  • 76) 0.135 874 531 328 × 2 = 0 + 0.271 749 062 656;
  • 77) 0.271 749 062 656 × 2 = 0 + 0.543 498 125 312;
  • 78) 0.543 498 125 312 × 2 = 1 + 0.086 996 250 624;
  • 79) 0.086 996 250 624 × 2 = 0 + 0.173 992 501 248;
  • 80) 0.173 992 501 248 × 2 = 0 + 0.347 985 002 496;
  • 81) 0.347 985 002 496 × 2 = 0 + 0.695 970 004 992;
  • 82) 0.695 970 004 992 × 2 = 1 + 0.391 940 009 984;
  • 83) 0.391 940 009 984 × 2 = 0 + 0.783 880 019 968;
  • 84) 0.783 880 019 968 × 2 = 1 + 0.567 760 039 936;
  • 85) 0.567 760 039 936 × 2 = 1 + 0.135 520 079 872;
  • 86) 0.135 520 079 872 × 2 = 0 + 0.271 040 159 744;
  • 87) 0.271 040 159 744 × 2 = 0 + 0.542 080 319 488;
  • 88) 0.542 080 319 488 × 2 = 1 + 0.084 160 638 976;
  • 89) 0.084 160 638 976 × 2 = 0 + 0.168 321 277 952;

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:


0.000 000 000 008 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0101 1000 1000 1001 1000 1100 1010 0000 0110 0100 0101 1001 0(2)

6. Positive number before normalization:

0.000 000 000 008 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0101 1000 1000 1001 1000 1100 1010 0000 0110 0100 0101 1001 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 008 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0101 1000 1000 1001 1000 1100 1010 0000 0110 0100 0101 1001 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0101 1000 1000 1001 1000 1100 1010 0000 0110 0100 0101 1001 0(2) × 20 =

1.0010 1011 0001 0001 0011 0001 1001 0100 0000 1100 1000 1011 0010(2) × 2-37


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): -37

Mantissa (not normalized):
1.0010 1011 0001 0001 0011 0001 1001 0100 0000 1100 1000 1011 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-37 + 2(11-1) - 1 =

(-37 + 1 023)(10) =

986(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;
  • 986 ÷ 2 = 493 + 0;
  • 493 ÷ 2 = 246 + 1;
  • 246 ÷ 2 = 123 + 0;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

986(10) =

011 1101 1010(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. 0010 1011 0001 0001 0011 0001 1001 0100 0000 1100 1000 1011 0010 =

0010 1011 0001 0001 0011 0001 1001 0100 0000 1100 1000 1011 0010


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 1101 1010

Mantissa (52 bits) =
0010 1011 0001 0001 0011 0001 1001 0100 0000 1100 1000 1011 0010


Decimal number -0.000 000 000 008 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 1010 - 0010 1011 0001 0001 0011 0001 1001 0100 0000 1100 1000 1011 0010

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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