6.285 714 285 714 285 689 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.285 714 285 714 285 689 9(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
6.285 714 285 714 285 689 9(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. First, convert to binary (in base 2) the integer part: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.

6(10) =

110(2)


3. Convert to binary (base 2) the fractional part: 0.285 714 285 714 285 689 9.

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) 0.285 714 285 714 285 689 9 × 2 = 0 + 0.571 428 571 428 571 379 8;
  • 2) 0.571 428 571 428 571 379 8 × 2 = 1 + 0.142 857 142 857 142 759 6;
  • 3) 0.142 857 142 857 142 759 6 × 2 = 0 + 0.285 714 285 714 285 519 2;
  • 4) 0.285 714 285 714 285 519 2 × 2 = 0 + 0.571 428 571 428 571 038 4;
  • 5) 0.571 428 571 428 571 038 4 × 2 = 1 + 0.142 857 142 857 142 076 8;
  • 6) 0.142 857 142 857 142 076 8 × 2 = 0 + 0.285 714 285 714 284 153 6;
  • 7) 0.285 714 285 714 284 153 6 × 2 = 0 + 0.571 428 571 428 568 307 2;
  • 8) 0.571 428 571 428 568 307 2 × 2 = 1 + 0.142 857 142 857 136 614 4;
  • 9) 0.142 857 142 857 136 614 4 × 2 = 0 + 0.285 714 285 714 273 228 8;
  • 10) 0.285 714 285 714 273 228 8 × 2 = 0 + 0.571 428 571 428 546 457 6;
  • 11) 0.571 428 571 428 546 457 6 × 2 = 1 + 0.142 857 142 857 092 915 2;
  • 12) 0.142 857 142 857 092 915 2 × 2 = 0 + 0.285 714 285 714 185 830 4;
  • 13) 0.285 714 285 714 185 830 4 × 2 = 0 + 0.571 428 571 428 371 660 8;
  • 14) 0.571 428 571 428 371 660 8 × 2 = 1 + 0.142 857 142 856 743 321 6;
  • 15) 0.142 857 142 856 743 321 6 × 2 = 0 + 0.285 714 285 713 486 643 2;
  • 16) 0.285 714 285 713 486 643 2 × 2 = 0 + 0.571 428 571 426 973 286 4;
  • 17) 0.571 428 571 426 973 286 4 × 2 = 1 + 0.142 857 142 853 946 572 8;
  • 18) 0.142 857 142 853 946 572 8 × 2 = 0 + 0.285 714 285 707 893 145 6;
  • 19) 0.285 714 285 707 893 145 6 × 2 = 0 + 0.571 428 571 415 786 291 2;
  • 20) 0.571 428 571 415 786 291 2 × 2 = 1 + 0.142 857 142 831 572 582 4;
  • 21) 0.142 857 142 831 572 582 4 × 2 = 0 + 0.285 714 285 663 145 164 8;
  • 22) 0.285 714 285 663 145 164 8 × 2 = 0 + 0.571 428 571 326 290 329 6;
  • 23) 0.571 428 571 326 290 329 6 × 2 = 1 + 0.142 857 142 652 580 659 2;
  • 24) 0.142 857 142 652 580 659 2 × 2 = 0 + 0.285 714 285 305 161 318 4;
  • 25) 0.285 714 285 305 161 318 4 × 2 = 0 + 0.571 428 570 610 322 636 8;
  • 26) 0.571 428 570 610 322 636 8 × 2 = 1 + 0.142 857 141 220 645 273 6;
  • 27) 0.142 857 141 220 645 273 6 × 2 = 0 + 0.285 714 282 441 290 547 2;
  • 28) 0.285 714 282 441 290 547 2 × 2 = 0 + 0.571 428 564 882 581 094 4;
  • 29) 0.571 428 564 882 581 094 4 × 2 = 1 + 0.142 857 129 765 162 188 8;
  • 30) 0.142 857 129 765 162 188 8 × 2 = 0 + 0.285 714 259 530 324 377 6;
  • 31) 0.285 714 259 530 324 377 6 × 2 = 0 + 0.571 428 519 060 648 755 2;
  • 32) 0.571 428 519 060 648 755 2 × 2 = 1 + 0.142 857 038 121 297 510 4;
  • 33) 0.142 857 038 121 297 510 4 × 2 = 0 + 0.285 714 076 242 595 020 8;
  • 34) 0.285 714 076 242 595 020 8 × 2 = 0 + 0.571 428 152 485 190 041 6;
  • 35) 0.571 428 152 485 190 041 6 × 2 = 1 + 0.142 856 304 970 380 083 2;
  • 36) 0.142 856 304 970 380 083 2 × 2 = 0 + 0.285 712 609 940 760 166 4;
  • 37) 0.285 712 609 940 760 166 4 × 2 = 0 + 0.571 425 219 881 520 332 8;
  • 38) 0.571 425 219 881 520 332 8 × 2 = 1 + 0.142 850 439 763 040 665 6;
  • 39) 0.142 850 439 763 040 665 6 × 2 = 0 + 0.285 700 879 526 081 331 2;
  • 40) 0.285 700 879 526 081 331 2 × 2 = 0 + 0.571 401 759 052 162 662 4;
  • 41) 0.571 401 759 052 162 662 4 × 2 = 1 + 0.142 803 518 104 325 324 8;
  • 42) 0.142 803 518 104 325 324 8 × 2 = 0 + 0.285 607 036 208 650 649 6;
  • 43) 0.285 607 036 208 650 649 6 × 2 = 0 + 0.571 214 072 417 301 299 2;
  • 44) 0.571 214 072 417 301 299 2 × 2 = 1 + 0.142 428 144 834 602 598 4;
  • 45) 0.142 428 144 834 602 598 4 × 2 = 0 + 0.284 856 289 669 205 196 8;
  • 46) 0.284 856 289 669 205 196 8 × 2 = 0 + 0.569 712 579 338 410 393 6;
  • 47) 0.569 712 579 338 410 393 6 × 2 = 1 + 0.139 425 158 676 820 787 2;
  • 48) 0.139 425 158 676 820 787 2 × 2 = 0 + 0.278 850 317 353 641 574 4;
  • 49) 0.278 850 317 353 641 574 4 × 2 = 0 + 0.557 700 634 707 283 148 8;
  • 50) 0.557 700 634 707 283 148 8 × 2 = 1 + 0.115 401 269 414 566 297 6;
  • 51) 0.115 401 269 414 566 297 6 × 2 = 0 + 0.230 802 538 829 132 595 2;
  • 52) 0.230 802 538 829 132 595 2 × 2 = 0 + 0.461 605 077 658 265 190 4;
  • 53) 0.461 605 077 658 265 190 4 × 2 = 0 + 0.923 210 155 316 530 380 8;

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).


4. 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:


0.285 714 285 714 285 689 9(10) =

0.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2)

5. Positive number before normalization:

6.285 714 285 714 285 689 9(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2)

6. 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:


6.285 714 285 714 285 689 9(10) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2) =

110.0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 0(2) × 20 =

1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 000(2) × 22


7. 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 0 (a positive number)

Exponent (unadjusted): 2

Mantissa (not normalized):
1.1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 000


8. 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)


9. 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;

10. 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)


11. 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. 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 000 =

1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0001

Mantissa (52 bits) =
1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001


Decimal number 6.285 714 285 714 285 689 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001 0010 0100 1001

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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