Convert -6.681 911 775 230 489 115 351 6 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

-6.681 911 775 230 489 115 351 6(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. Start with the positive version of the number:

|-6.681 911 775 230 489 115 351 6| = 6.681 911 775 230 489 115 351 6

2. First, convert to the binary (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;

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.

6(10) =


110(2)


4. Convert to the binary (base 2) the fractional part: 0.681 911 775 230 489 115 351 6.

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) 0.681 911 775 230 489 115 351 6 × 2 = 1 + 0.363 823 550 460 978 230 703 2;
  • 2) 0.363 823 550 460 978 230 703 2 × 2 = 0 + 0.727 647 100 921 956 461 406 4;
  • 3) 0.727 647 100 921 956 461 406 4 × 2 = 1 + 0.455 294 201 843 912 922 812 8;
  • 4) 0.455 294 201 843 912 922 812 8 × 2 = 0 + 0.910 588 403 687 825 845 625 6;
  • 5) 0.910 588 403 687 825 845 625 6 × 2 = 1 + 0.821 176 807 375 651 691 251 2;
  • 6) 0.821 176 807 375 651 691 251 2 × 2 = 1 + 0.642 353 614 751 303 382 502 4;
  • 7) 0.642 353 614 751 303 382 502 4 × 2 = 1 + 0.284 707 229 502 606 765 004 8;
  • 8) 0.284 707 229 502 606 765 004 8 × 2 = 0 + 0.569 414 459 005 213 530 009 6;
  • 9) 0.569 414 459 005 213 530 009 6 × 2 = 1 + 0.138 828 918 010 427 060 019 2;
  • 10) 0.138 828 918 010 427 060 019 2 × 2 = 0 + 0.277 657 836 020 854 120 038 4;
  • 11) 0.277 657 836 020 854 120 038 4 × 2 = 0 + 0.555 315 672 041 708 240 076 8;
  • 12) 0.555 315 672 041 708 240 076 8 × 2 = 1 + 0.110 631 344 083 416 480 153 6;
  • 13) 0.110 631 344 083 416 480 153 6 × 2 = 0 + 0.221 262 688 166 832 960 307 2;
  • 14) 0.221 262 688 166 832 960 307 2 × 2 = 0 + 0.442 525 376 333 665 920 614 4;
  • 15) 0.442 525 376 333 665 920 614 4 × 2 = 0 + 0.885 050 752 667 331 841 228 8;
  • 16) 0.885 050 752 667 331 841 228 8 × 2 = 1 + 0.770 101 505 334 663 682 457 6;
  • 17) 0.770 101 505 334 663 682 457 6 × 2 = 1 + 0.540 203 010 669 327 364 915 2;
  • 18) 0.540 203 010 669 327 364 915 2 × 2 = 1 + 0.080 406 021 338 654 729 830 4;
  • 19) 0.080 406 021 338 654 729 830 4 × 2 = 0 + 0.160 812 042 677 309 459 660 8;
  • 20) 0.160 812 042 677 309 459 660 8 × 2 = 0 + 0.321 624 085 354 618 919 321 6;
  • 21) 0.321 624 085 354 618 919 321 6 × 2 = 0 + 0.643 248 170 709 237 838 643 2;
  • 22) 0.643 248 170 709 237 838 643 2 × 2 = 1 + 0.286 496 341 418 475 677 286 4;
  • 23) 0.286 496 341 418 475 677 286 4 × 2 = 0 + 0.572 992 682 836 951 354 572 8;
  • 24) 0.572 992 682 836 951 354 572 8 × 2 = 1 + 0.145 985 365 673 902 709 145 6;
  • 25) 0.145 985 365 673 902 709 145 6 × 2 = 0 + 0.291 970 731 347 805 418 291 2;
  • 26) 0.291 970 731 347 805 418 291 2 × 2 = 0 + 0.583 941 462 695 610 836 582 4;
  • 27) 0.583 941 462 695 610 836 582 4 × 2 = 1 + 0.167 882 925 391 221 673 164 8;
  • 28) 0.167 882 925 391 221 673 164 8 × 2 = 0 + 0.335 765 850 782 443 346 329 6;
  • 29) 0.335 765 850 782 443 346 329 6 × 2 = 0 + 0.671 531 701 564 886 692 659 2;
  • 30) 0.671 531 701 564 886 692 659 2 × 2 = 1 + 0.343 063 403 129 773 385 318 4;
  • 31) 0.343 063 403 129 773 385 318 4 × 2 = 0 + 0.686 126 806 259 546 770 636 8;
  • 32) 0.686 126 806 259 546 770 636 8 × 2 = 1 + 0.372 253 612 519 093 541 273 6;
  • 33) 0.372 253 612 519 093 541 273 6 × 2 = 0 + 0.744 507 225 038 187 082 547 2;
  • 34) 0.744 507 225 038 187 082 547 2 × 2 = 1 + 0.489 014 450 076 374 165 094 4;
  • 35) 0.489 014 450 076 374 165 094 4 × 2 = 0 + 0.978 028 900 152 748 330 188 8;
  • 36) 0.978 028 900 152 748 330 188 8 × 2 = 1 + 0.956 057 800 305 496 660 377 6;
  • 37) 0.956 057 800 305 496 660 377 6 × 2 = 1 + 0.912 115 600 610 993 320 755 2;
  • 38) 0.912 115 600 610 993 320 755 2 × 2 = 1 + 0.824 231 201 221 986 641 510 4;
  • 39) 0.824 231 201 221 986 641 510 4 × 2 = 1 + 0.648 462 402 443 973 283 020 8;
  • 40) 0.648 462 402 443 973 283 020 8 × 2 = 1 + 0.296 924 804 887 946 566 041 6;
  • 41) 0.296 924 804 887 946 566 041 6 × 2 = 0 + 0.593 849 609 775 893 132 083 2;
  • 42) 0.593 849 609 775 893 132 083 2 × 2 = 1 + 0.187 699 219 551 786 264 166 4;
  • 43) 0.187 699 219 551 786 264 166 4 × 2 = 0 + 0.375 398 439 103 572 528 332 8;
  • 44) 0.375 398 439 103 572 528 332 8 × 2 = 0 + 0.750 796 878 207 145 056 665 6;
  • 45) 0.750 796 878 207 145 056 665 6 × 2 = 1 + 0.501 593 756 414 290 113 331 2;
  • 46) 0.501 593 756 414 290 113 331 2 × 2 = 1 + 0.003 187 512 828 580 226 662 4;
  • 47) 0.003 187 512 828 580 226 662 4 × 2 = 0 + 0.006 375 025 657 160 453 324 8;
  • 48) 0.006 375 025 657 160 453 324 8 × 2 = 0 + 0.012 750 051 314 320 906 649 6;
  • 49) 0.012 750 051 314 320 906 649 6 × 2 = 0 + 0.025 500 102 628 641 813 299 2;
  • 50) 0.025 500 102 628 641 813 299 2 × 2 = 0 + 0.051 000 205 257 283 626 598 4;
  • 51) 0.051 000 205 257 283 626 598 4 × 2 = 0 + 0.102 000 410 514 567 253 196 8;
  • 52) 0.102 000 410 514 567 253 196 8 × 2 = 0 + 0.204 000 821 029 134 506 393 6;
  • 53) 0.204 000 821 029 134 506 393 6 × 2 = 0 + 0.408 001 642 058 269 012 787 2;

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...)


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:

0.681 911 775 230 489 115 351 6(10) =


0.1010 1110 1001 0001 1100 0101 0010 0101 0101 1111 0100 1100 0000 0(2)


6. Positive number before normalization:

6.681 911 775 230 489 115 351 6(10) =


110.1010 1110 1001 0001 1100 0101 0010 0101 0101 1111 0100 1100 0000 0(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:

6.681 911 775 230 489 115 351 6(10) =


110.1010 1110 1001 0001 1100 0101 0010 0101 0101 1111 0100 1100 0000 0(2) =


110.1010 1110 1001 0001 1100 0101 0010 0101 0101 1111 0100 1100 0000 0(2) × 20 =


1.1010 1011 1010 0100 0111 0001 0100 1001 0101 0111 1101 0011 0000 000(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):
1.1010 1011 1010 0100 0111 0001 0100 1001 0101 0111 1101 0011 0000 000


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. 1010 1011 1010 0100 0111 0001 0100 1001 0101 0111 1101 0011 0000 000 =


1010 1011 1010 0100 0111 0001 0100 1001 0101 0111 1101 0011 0000


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) =
1010 1011 1010 0100 0111 0001 0100 1001 0101 0111 1101 0011 0000


Number -6.681 911 775 230 489 115 351 6 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
1 - 100 0000 0001 - 1010 1011 1010 0100 0111 0001 0100 1001 0101 0111 1101 0011 0000

(64 bits IEEE 754)
  • Sign (1 bit):

    • 1

      63
  • Exponent (11 bits):

    • 1

      62
    • 0

      61
    • 0

      60
    • 0

      59
    • 0

      58
    • 0

      57
    • 0

      56
    • 0

      55
    • 0

      54
    • 0

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 1

      49
    • 0

      48
    • 1

      47
    • 0

      46
    • 1

      45
    • 1

      44
    • 1

      43
    • 0

      42
    • 1

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 0

      37
    • 0

      36
    • 0

      35
    • 1

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 0

      29
    • 1

      28
    • 0

      27
    • 1

      26
    • 0

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 1

      16
    • 0

      15
    • 1

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 0

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 0

      3
    • 0

      2
    • 0

      1
    • 0

      0

More operations of this kind:

-6.681 911 775 230 489 115 351 7 = ? ... -6.681 911 775 230 489 115 351 5 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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