-7.769 489 691 003 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -7.769 489 691 003 53(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
-7.769 489 691 003 53(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:

|-7.769 489 691 003 53| = 7.769 489 691 003 53


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: 0.769 489 691 003 53.

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) 0.769 489 691 003 53 × 2 = 1 + 0.538 979 382 007 06;
  • 2) 0.538 979 382 007 06 × 2 = 1 + 0.077 958 764 014 12;
  • 3) 0.077 958 764 014 12 × 2 = 0 + 0.155 917 528 028 24;
  • 4) 0.155 917 528 028 24 × 2 = 0 + 0.311 835 056 056 48;
  • 5) 0.311 835 056 056 48 × 2 = 0 + 0.623 670 112 112 96;
  • 6) 0.623 670 112 112 96 × 2 = 1 + 0.247 340 224 225 92;
  • 7) 0.247 340 224 225 92 × 2 = 0 + 0.494 680 448 451 84;
  • 8) 0.494 680 448 451 84 × 2 = 0 + 0.989 360 896 903 68;
  • 9) 0.989 360 896 903 68 × 2 = 1 + 0.978 721 793 807 36;
  • 10) 0.978 721 793 807 36 × 2 = 1 + 0.957 443 587 614 72;
  • 11) 0.957 443 587 614 72 × 2 = 1 + 0.914 887 175 229 44;
  • 12) 0.914 887 175 229 44 × 2 = 1 + 0.829 774 350 458 88;
  • 13) 0.829 774 350 458 88 × 2 = 1 + 0.659 548 700 917 76;
  • 14) 0.659 548 700 917 76 × 2 = 1 + 0.319 097 401 835 52;
  • 15) 0.319 097 401 835 52 × 2 = 0 + 0.638 194 803 671 04;
  • 16) 0.638 194 803 671 04 × 2 = 1 + 0.276 389 607 342 08;
  • 17) 0.276 389 607 342 08 × 2 = 0 + 0.552 779 214 684 16;
  • 18) 0.552 779 214 684 16 × 2 = 1 + 0.105 558 429 368 32;
  • 19) 0.105 558 429 368 32 × 2 = 0 + 0.211 116 858 736 64;
  • 20) 0.211 116 858 736 64 × 2 = 0 + 0.422 233 717 473 28;
  • 21) 0.422 233 717 473 28 × 2 = 0 + 0.844 467 434 946 56;
  • 22) 0.844 467 434 946 56 × 2 = 1 + 0.688 934 869 893 12;
  • 23) 0.688 934 869 893 12 × 2 = 1 + 0.377 869 739 786 24;
  • 24) 0.377 869 739 786 24 × 2 = 0 + 0.755 739 479 572 48;
  • 25) 0.755 739 479 572 48 × 2 = 1 + 0.511 478 959 144 96;
  • 26) 0.511 478 959 144 96 × 2 = 1 + 0.022 957 918 289 92;
  • 27) 0.022 957 918 289 92 × 2 = 0 + 0.045 915 836 579 84;
  • 28) 0.045 915 836 579 84 × 2 = 0 + 0.091 831 673 159 68;
  • 29) 0.091 831 673 159 68 × 2 = 0 + 0.183 663 346 319 36;
  • 30) 0.183 663 346 319 36 × 2 = 0 + 0.367 326 692 638 72;
  • 31) 0.367 326 692 638 72 × 2 = 0 + 0.734 653 385 277 44;
  • 32) 0.734 653 385 277 44 × 2 = 1 + 0.469 306 770 554 88;
  • 33) 0.469 306 770 554 88 × 2 = 0 + 0.938 613 541 109 76;
  • 34) 0.938 613 541 109 76 × 2 = 1 + 0.877 227 082 219 52;
  • 35) 0.877 227 082 219 52 × 2 = 1 + 0.754 454 164 439 04;
  • 36) 0.754 454 164 439 04 × 2 = 1 + 0.508 908 328 878 08;
  • 37) 0.508 908 328 878 08 × 2 = 1 + 0.017 816 657 756 16;
  • 38) 0.017 816 657 756 16 × 2 = 0 + 0.035 633 315 512 32;
  • 39) 0.035 633 315 512 32 × 2 = 0 + 0.071 266 631 024 64;
  • 40) 0.071 266 631 024 64 × 2 = 0 + 0.142 533 262 049 28;
  • 41) 0.142 533 262 049 28 × 2 = 0 + 0.285 066 524 098 56;
  • 42) 0.285 066 524 098 56 × 2 = 0 + 0.570 133 048 197 12;
  • 43) 0.570 133 048 197 12 × 2 = 1 + 0.140 266 096 394 24;
  • 44) 0.140 266 096 394 24 × 2 = 0 + 0.280 532 192 788 48;
  • 45) 0.280 532 192 788 48 × 2 = 0 + 0.561 064 385 576 96;
  • 46) 0.561 064 385 576 96 × 2 = 1 + 0.122 128 771 153 92;
  • 47) 0.122 128 771 153 92 × 2 = 0 + 0.244 257 542 307 84;
  • 48) 0.244 257 542 307 84 × 2 = 0 + 0.488 515 084 615 68;
  • 49) 0.488 515 084 615 68 × 2 = 0 + 0.977 030 169 231 36;
  • 50) 0.977 030 169 231 36 × 2 = 1 + 0.954 060 338 462 72;
  • 51) 0.954 060 338 462 72 × 2 = 1 + 0.908 120 676 925 44;
  • 52) 0.908 120 676 925 44 × 2 = 1 + 0.816 241 353 850 88;
  • 53) 0.816 241 353 850 88 × 2 = 1 + 0.632 482 707 701 76;

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:


0.769 489 691 003 53(10) =

0.1100 0100 1111 1101 0100 0110 1100 0001 0111 1000 0010 0100 0111 1(2)

6. Positive number before normalization:

7.769 489 691 003 53(10) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 1000 0010 0100 0111 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:


7.769 489 691 003 53(10) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 1000 0010 0100 0111 1(2) =

111.1100 0100 1111 1101 0100 0110 1100 0001 0111 1000 0010 0100 0111 1(2) × 20 =

1.1111 0001 0011 1111 0101 0001 1011 0000 0101 1110 0000 1001 0001 111(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 1110 0000 1001 0001 111


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 1110 0000 1001 0001 111 =

1111 0001 0011 1111 0101 0001 1011 0000 0101 1110 0000 1001 0001


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 1110 0000 1001 0001


Decimal number -7.769 489 691 003 53 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 1110 0000 1001 0001

» Convert decimal -7.769 489 691 004 01 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