-7.769 489 691 002 67 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -7.769 489 691 002 67(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
-7.769 489 691 002 67(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:

|-7.769 489 691 002 67| = 7.769 489 691 002 67


2. First, convert to binary (in base 2) the integer part: 7.
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;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

7(10) =

111(2)


4. Convert to binary (base 2) the fractional part: 0.769 489 691 002 67.

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) 0.769 489 691 002 67 × 2 = 1 + 0.538 979 382 005 34;
  • 2) 0.538 979 382 005 34 × 2 = 1 + 0.077 958 764 010 68;
  • 3) 0.077 958 764 010 68 × 2 = 0 + 0.155 917 528 021 36;
  • 4) 0.155 917 528 021 36 × 2 = 0 + 0.311 835 056 042 72;
  • 5) 0.311 835 056 042 72 × 2 = 0 + 0.623 670 112 085 44;
  • 6) 0.623 670 112 085 44 × 2 = 1 + 0.247 340 224 170 88;
  • 7) 0.247 340 224 170 88 × 2 = 0 + 0.494 680 448 341 76;
  • 8) 0.494 680 448 341 76 × 2 = 0 + 0.989 360 896 683 52;
  • 9) 0.989 360 896 683 52 × 2 = 1 + 0.978 721 793 367 04;
  • 10) 0.978 721 793 367 04 × 2 = 1 + 0.957 443 586 734 08;
  • 11) 0.957 443 586 734 08 × 2 = 1 + 0.914 887 173 468 16;
  • 12) 0.914 887 173 468 16 × 2 = 1 + 0.829 774 346 936 32;
  • 13) 0.829 774 346 936 32 × 2 = 1 + 0.659 548 693 872 64;
  • 14) 0.659 548 693 872 64 × 2 = 1 + 0.319 097 387 745 28;
  • 15) 0.319 097 387 745 28 × 2 = 0 + 0.638 194 775 490 56;
  • 16) 0.638 194 775 490 56 × 2 = 1 + 0.276 389 550 981 12;
  • 17) 0.276 389 550 981 12 × 2 = 0 + 0.552 779 101 962 24;
  • 18) 0.552 779 101 962 24 × 2 = 1 + 0.105 558 203 924 48;
  • 19) 0.105 558 203 924 48 × 2 = 0 + 0.211 116 407 848 96;
  • 20) 0.211 116 407 848 96 × 2 = 0 + 0.422 232 815 697 92;
  • 21) 0.422 232 815 697 92 × 2 = 0 + 0.844 465 631 395 84;
  • 22) 0.844 465 631 395 84 × 2 = 1 + 0.688 931 262 791 68;
  • 23) 0.688 931 262 791 68 × 2 = 1 + 0.377 862 525 583 36;
  • 24) 0.377 862 525 583 36 × 2 = 0 + 0.755 725 051 166 72;
  • 25) 0.755 725 051 166 72 × 2 = 1 + 0.511 450 102 333 44;
  • 26) 0.511 450 102 333 44 × 2 = 1 + 0.022 900 204 666 88;
  • 27) 0.022 900 204 666 88 × 2 = 0 + 0.045 800 409 333 76;
  • 28) 0.045 800 409 333 76 × 2 = 0 + 0.091 600 818 667 52;
  • 29) 0.091 600 818 667 52 × 2 = 0 + 0.183 201 637 335 04;
  • 30) 0.183 201 637 335 04 × 2 = 0 + 0.366 403 274 670 08;
  • 31) 0.366 403 274 670 08 × 2 = 0 + 0.732 806 549 340 16;
  • 32) 0.732 806 549 340 16 × 2 = 1 + 0.465 613 098 680 32;
  • 33) 0.465 613 098 680 32 × 2 = 0 + 0.931 226 197 360 64;
  • 34) 0.931 226 197 360 64 × 2 = 1 + 0.862 452 394 721 28;
  • 35) 0.862 452 394 721 28 × 2 = 1 + 0.724 904 789 442 56;
  • 36) 0.724 904 789 442 56 × 2 = 1 + 0.449 809 578 885 12;
  • 37) 0.449 809 578 885 12 × 2 = 0 + 0.899 619 157 770 24;
  • 38) 0.899 619 157 770 24 × 2 = 1 + 0.799 238 315 540 48;
  • 39) 0.799 238 315 540 48 × 2 = 1 + 0.598 476 631 080 96;
  • 40) 0.598 476 631 080 96 × 2 = 1 + 0.196 953 262 161 92;
  • 41) 0.196 953 262 161 92 × 2 = 0 + 0.393 906 524 323 84;
  • 42) 0.393 906 524 323 84 × 2 = 0 + 0.787 813 048 647 68;
  • 43) 0.787 813 048 647 68 × 2 = 1 + 0.575 626 097 295 36;
  • 44) 0.575 626 097 295 36 × 2 = 1 + 0.151 252 194 590 72;
  • 45) 0.151 252 194 590 72 × 2 = 0 + 0.302 504 389 181 44;
  • 46) 0.302 504 389 181 44 × 2 = 0 + 0.605 008 778 362 88;
  • 47) 0.605 008 778 362 88 × 2 = 1 + 0.210 017 556 725 76;
  • 48) 0.210 017 556 725 76 × 2 = 0 + 0.420 035 113 451 52;
  • 49) 0.420 035 113 451 52 × 2 = 0 + 0.840 070 226 903 04;
  • 50) 0.840 070 226 903 04 × 2 = 1 + 0.680 140 453 806 08;
  • 51) 0.680 140 453 806 08 × 2 = 1 + 0.360 280 907 612 16;
  • 52) 0.360 280 907 612 16 × 2 = 0 + 0.720 561 815 224 32;
  • 53) 0.720 561 815 224 32 × 2 = 1 + 0.441 123 630 448 64;

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:


0.769 489 691 002 67(10) =

0.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0011 0010 0110 1(2)

6. Positive number before normalization:

7.769 489 691 002 67(10) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0011 0010 0110 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


7.769 489 691 002 67(10) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0011 0010 0110 1(2) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 0111 0011 0010 0110 1(2) × 20 =

1.1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1100 1001 101(2) × 22


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): 2

Mantissa (not normalized):
1.1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1100 1001 101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1025(10) =

100 0000 0001(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1100 1001 101 =

1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1100 1001


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) =
100 0000 0001

Mantissa (52 bits) =
1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1100 1001


Decimal number -7.769 489 691 002 67 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0001 - 1111 0001 0011 1111 0101 0001 1011 0000 0101 1101 1100 1100 1001

» Convert decimal -7.769 489 691 002 12 to 64 bit double precision IEEE 754 binary floating point representation standard

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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