-0.000 375 658 489 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 489 32(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
-0.000 375 658 489 32(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:

|-0.000 375 658 489 32| = 0.000 375 658 489 32


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: 0.000 375 658 489 32.

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) 0.000 375 658 489 32 × 2 = 0 + 0.000 751 316 978 64;
  • 2) 0.000 751 316 978 64 × 2 = 0 + 0.001 502 633 957 28;
  • 3) 0.001 502 633 957 28 × 2 = 0 + 0.003 005 267 914 56;
  • 4) 0.003 005 267 914 56 × 2 = 0 + 0.006 010 535 829 12;
  • 5) 0.006 010 535 829 12 × 2 = 0 + 0.012 021 071 658 24;
  • 6) 0.012 021 071 658 24 × 2 = 0 + 0.024 042 143 316 48;
  • 7) 0.024 042 143 316 48 × 2 = 0 + 0.048 084 286 632 96;
  • 8) 0.048 084 286 632 96 × 2 = 0 + 0.096 168 573 265 92;
  • 9) 0.096 168 573 265 92 × 2 = 0 + 0.192 337 146 531 84;
  • 10) 0.192 337 146 531 84 × 2 = 0 + 0.384 674 293 063 68;
  • 11) 0.384 674 293 063 68 × 2 = 0 + 0.769 348 586 127 36;
  • 12) 0.769 348 586 127 36 × 2 = 1 + 0.538 697 172 254 72;
  • 13) 0.538 697 172 254 72 × 2 = 1 + 0.077 394 344 509 44;
  • 14) 0.077 394 344 509 44 × 2 = 0 + 0.154 788 689 018 88;
  • 15) 0.154 788 689 018 88 × 2 = 0 + 0.309 577 378 037 76;
  • 16) 0.309 577 378 037 76 × 2 = 0 + 0.619 154 756 075 52;
  • 17) 0.619 154 756 075 52 × 2 = 1 + 0.238 309 512 151 04;
  • 18) 0.238 309 512 151 04 × 2 = 0 + 0.476 619 024 302 08;
  • 19) 0.476 619 024 302 08 × 2 = 0 + 0.953 238 048 604 16;
  • 20) 0.953 238 048 604 16 × 2 = 1 + 0.906 476 097 208 32;
  • 21) 0.906 476 097 208 32 × 2 = 1 + 0.812 952 194 416 64;
  • 22) 0.812 952 194 416 64 × 2 = 1 + 0.625 904 388 833 28;
  • 23) 0.625 904 388 833 28 × 2 = 1 + 0.251 808 777 666 56;
  • 24) 0.251 808 777 666 56 × 2 = 0 + 0.503 617 555 333 12;
  • 25) 0.503 617 555 333 12 × 2 = 1 + 0.007 235 110 666 24;
  • 26) 0.007 235 110 666 24 × 2 = 0 + 0.014 470 221 332 48;
  • 27) 0.014 470 221 332 48 × 2 = 0 + 0.028 940 442 664 96;
  • 28) 0.028 940 442 664 96 × 2 = 0 + 0.057 880 885 329 92;
  • 29) 0.057 880 885 329 92 × 2 = 0 + 0.115 761 770 659 84;
  • 30) 0.115 761 770 659 84 × 2 = 0 + 0.231 523 541 319 68;
  • 31) 0.231 523 541 319 68 × 2 = 0 + 0.463 047 082 639 36;
  • 32) 0.463 047 082 639 36 × 2 = 0 + 0.926 094 165 278 72;
  • 33) 0.926 094 165 278 72 × 2 = 1 + 0.852 188 330 557 44;
  • 34) 0.852 188 330 557 44 × 2 = 1 + 0.704 376 661 114 88;
  • 35) 0.704 376 661 114 88 × 2 = 1 + 0.408 753 322 229 76;
  • 36) 0.408 753 322 229 76 × 2 = 0 + 0.817 506 644 459 52;
  • 37) 0.817 506 644 459 52 × 2 = 1 + 0.635 013 288 919 04;
  • 38) 0.635 013 288 919 04 × 2 = 1 + 0.270 026 577 838 08;
  • 39) 0.270 026 577 838 08 × 2 = 0 + 0.540 053 155 676 16;
  • 40) 0.540 053 155 676 16 × 2 = 1 + 0.080 106 311 352 32;
  • 41) 0.080 106 311 352 32 × 2 = 0 + 0.160 212 622 704 64;
  • 42) 0.160 212 622 704 64 × 2 = 0 + 0.320 425 245 409 28;
  • 43) 0.320 425 245 409 28 × 2 = 0 + 0.640 850 490 818 56;
  • 44) 0.640 850 490 818 56 × 2 = 1 + 0.281 700 981 637 12;
  • 45) 0.281 700 981 637 12 × 2 = 0 + 0.563 401 963 274 24;
  • 46) 0.563 401 963 274 24 × 2 = 1 + 0.126 803 926 548 48;
  • 47) 0.126 803 926 548 48 × 2 = 0 + 0.253 607 853 096 96;
  • 48) 0.253 607 853 096 96 × 2 = 0 + 0.507 215 706 193 92;
  • 49) 0.507 215 706 193 92 × 2 = 1 + 0.014 431 412 387 84;
  • 50) 0.014 431 412 387 84 × 2 = 0 + 0.028 862 824 775 68;
  • 51) 0.028 862 824 775 68 × 2 = 0 + 0.057 725 649 551 36;
  • 52) 0.057 725 649 551 36 × 2 = 0 + 0.115 451 299 102 72;
  • 53) 0.115 451 299 102 72 × 2 = 0 + 0.230 902 598 205 44;
  • 54) 0.230 902 598 205 44 × 2 = 0 + 0.461 805 196 410 88;
  • 55) 0.461 805 196 410 88 × 2 = 0 + 0.923 610 392 821 76;
  • 56) 0.923 610 392 821 76 × 2 = 1 + 0.847 220 785 643 52;
  • 57) 0.847 220 785 643 52 × 2 = 1 + 0.694 441 571 287 04;
  • 58) 0.694 441 571 287 04 × 2 = 1 + 0.388 883 142 574 08;
  • 59) 0.388 883 142 574 08 × 2 = 0 + 0.777 766 285 148 16;
  • 60) 0.777 766 285 148 16 × 2 = 1 + 0.555 532 570 296 32;
  • 61) 0.555 532 570 296 32 × 2 = 1 + 0.111 065 140 592 64;
  • 62) 0.111 065 140 592 64 × 2 = 0 + 0.222 130 281 185 28;
  • 63) 0.222 130 281 185 28 × 2 = 0 + 0.444 260 562 370 56;
  • 64) 0.444 260 562 370 56 × 2 = 0 + 0.888 521 124 741 12;

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:


0.000 375 658 489 32(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000(2)

6. Positive number before normalization:

0.000 375 658 489 32(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 375 658 489 32(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000(2) × 2-12


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): -12

Mantissa (not normalized):
1.1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000 =

1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000


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 0011

Mantissa (52 bits) =
1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000


Decimal number -0.000 375 658 489 32 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1101 0001 0100 1000 0001 1101 1000

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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