-0.000 164 779 749 346 824 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 164 779 749 346 824 7(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
-0.000 164 779 749 346 824 7(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:

|-0.000 164 779 749 346 824 7| = 0.000 164 779 749 346 824 7


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: 0.000 164 779 749 346 824 7.

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) 0.000 164 779 749 346 824 7 × 2 = 0 + 0.000 329 559 498 693 649 4;
  • 2) 0.000 329 559 498 693 649 4 × 2 = 0 + 0.000 659 118 997 387 298 8;
  • 3) 0.000 659 118 997 387 298 8 × 2 = 0 + 0.001 318 237 994 774 597 6;
  • 4) 0.001 318 237 994 774 597 6 × 2 = 0 + 0.002 636 475 989 549 195 2;
  • 5) 0.002 636 475 989 549 195 2 × 2 = 0 + 0.005 272 951 979 098 390 4;
  • 6) 0.005 272 951 979 098 390 4 × 2 = 0 + 0.010 545 903 958 196 780 8;
  • 7) 0.010 545 903 958 196 780 8 × 2 = 0 + 0.021 091 807 916 393 561 6;
  • 8) 0.021 091 807 916 393 561 6 × 2 = 0 + 0.042 183 615 832 787 123 2;
  • 9) 0.042 183 615 832 787 123 2 × 2 = 0 + 0.084 367 231 665 574 246 4;
  • 10) 0.084 367 231 665 574 246 4 × 2 = 0 + 0.168 734 463 331 148 492 8;
  • 11) 0.168 734 463 331 148 492 8 × 2 = 0 + 0.337 468 926 662 296 985 6;
  • 12) 0.337 468 926 662 296 985 6 × 2 = 0 + 0.674 937 853 324 593 971 2;
  • 13) 0.674 937 853 324 593 971 2 × 2 = 1 + 0.349 875 706 649 187 942 4;
  • 14) 0.349 875 706 649 187 942 4 × 2 = 0 + 0.699 751 413 298 375 884 8;
  • 15) 0.699 751 413 298 375 884 8 × 2 = 1 + 0.399 502 826 596 751 769 6;
  • 16) 0.399 502 826 596 751 769 6 × 2 = 0 + 0.799 005 653 193 503 539 2;
  • 17) 0.799 005 653 193 503 539 2 × 2 = 1 + 0.598 011 306 387 007 078 4;
  • 18) 0.598 011 306 387 007 078 4 × 2 = 1 + 0.196 022 612 774 014 156 8;
  • 19) 0.196 022 612 774 014 156 8 × 2 = 0 + 0.392 045 225 548 028 313 6;
  • 20) 0.392 045 225 548 028 313 6 × 2 = 0 + 0.784 090 451 096 056 627 2;
  • 21) 0.784 090 451 096 056 627 2 × 2 = 1 + 0.568 180 902 192 113 254 4;
  • 22) 0.568 180 902 192 113 254 4 × 2 = 1 + 0.136 361 804 384 226 508 8;
  • 23) 0.136 361 804 384 226 508 8 × 2 = 0 + 0.272 723 608 768 453 017 6;
  • 24) 0.272 723 608 768 453 017 6 × 2 = 0 + 0.545 447 217 536 906 035 2;
  • 25) 0.545 447 217 536 906 035 2 × 2 = 1 + 0.090 894 435 073 812 070 4;
  • 26) 0.090 894 435 073 812 070 4 × 2 = 0 + 0.181 788 870 147 624 140 8;
  • 27) 0.181 788 870 147 624 140 8 × 2 = 0 + 0.363 577 740 295 248 281 6;
  • 28) 0.363 577 740 295 248 281 6 × 2 = 0 + 0.727 155 480 590 496 563 2;
  • 29) 0.727 155 480 590 496 563 2 × 2 = 1 + 0.454 310 961 180 993 126 4;
  • 30) 0.454 310 961 180 993 126 4 × 2 = 0 + 0.908 621 922 361 986 252 8;
  • 31) 0.908 621 922 361 986 252 8 × 2 = 1 + 0.817 243 844 723 972 505 6;
  • 32) 0.817 243 844 723 972 505 6 × 2 = 1 + 0.634 487 689 447 945 011 2;
  • 33) 0.634 487 689 447 945 011 2 × 2 = 1 + 0.268 975 378 895 890 022 4;
  • 34) 0.268 975 378 895 890 022 4 × 2 = 0 + 0.537 950 757 791 780 044 8;
  • 35) 0.537 950 757 791 780 044 8 × 2 = 1 + 0.075 901 515 583 560 089 6;
  • 36) 0.075 901 515 583 560 089 6 × 2 = 0 + 0.151 803 031 167 120 179 2;
  • 37) 0.151 803 031 167 120 179 2 × 2 = 0 + 0.303 606 062 334 240 358 4;
  • 38) 0.303 606 062 334 240 358 4 × 2 = 0 + 0.607 212 124 668 480 716 8;
  • 39) 0.607 212 124 668 480 716 8 × 2 = 1 + 0.214 424 249 336 961 433 6;
  • 40) 0.214 424 249 336 961 433 6 × 2 = 0 + 0.428 848 498 673 922 867 2;
  • 41) 0.428 848 498 673 922 867 2 × 2 = 0 + 0.857 696 997 347 845 734 4;
  • 42) 0.857 696 997 347 845 734 4 × 2 = 1 + 0.715 393 994 695 691 468 8;
  • 43) 0.715 393 994 695 691 468 8 × 2 = 1 + 0.430 787 989 391 382 937 6;
  • 44) 0.430 787 989 391 382 937 6 × 2 = 0 + 0.861 575 978 782 765 875 2;
  • 45) 0.861 575 978 782 765 875 2 × 2 = 1 + 0.723 151 957 565 531 750 4;
  • 46) 0.723 151 957 565 531 750 4 × 2 = 1 + 0.446 303 915 131 063 500 8;
  • 47) 0.446 303 915 131 063 500 8 × 2 = 0 + 0.892 607 830 262 127 001 6;
  • 48) 0.892 607 830 262 127 001 6 × 2 = 1 + 0.785 215 660 524 254 003 2;
  • 49) 0.785 215 660 524 254 003 2 × 2 = 1 + 0.570 431 321 048 508 006 4;
  • 50) 0.570 431 321 048 508 006 4 × 2 = 1 + 0.140 862 642 097 016 012 8;
  • 51) 0.140 862 642 097 016 012 8 × 2 = 0 + 0.281 725 284 194 032 025 6;
  • 52) 0.281 725 284 194 032 025 6 × 2 = 0 + 0.563 450 568 388 064 051 2;
  • 53) 0.563 450 568 388 064 051 2 × 2 = 1 + 0.126 901 136 776 128 102 4;
  • 54) 0.126 901 136 776 128 102 4 × 2 = 0 + 0.253 802 273 552 256 204 8;
  • 55) 0.253 802 273 552 256 204 8 × 2 = 0 + 0.507 604 547 104 512 409 6;
  • 56) 0.507 604 547 104 512 409 6 × 2 = 1 + 0.015 209 094 209 024 819 2;
  • 57) 0.015 209 094 209 024 819 2 × 2 = 0 + 0.030 418 188 418 049 638 4;
  • 58) 0.030 418 188 418 049 638 4 × 2 = 0 + 0.060 836 376 836 099 276 8;
  • 59) 0.060 836 376 836 099 276 8 × 2 = 0 + 0.121 672 753 672 198 553 6;
  • 60) 0.121 672 753 672 198 553 6 × 2 = 0 + 0.243 345 507 344 397 107 2;
  • 61) 0.243 345 507 344 397 107 2 × 2 = 0 + 0.486 691 014 688 794 214 4;
  • 62) 0.486 691 014 688 794 214 4 × 2 = 0 + 0.973 382 029 377 588 428 8;
  • 63) 0.973 382 029 377 588 428 8 × 2 = 1 + 0.946 764 058 755 176 857 6;
  • 64) 0.946 764 058 755 176 857 6 × 2 = 1 + 0.893 528 117 510 353 715 2;
  • 65) 0.893 528 117 510 353 715 2 × 2 = 1 + 0.787 056 235 020 707 430 4;

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:


0.000 164 779 749 346 824 7(10) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1001 0000 0011 1(2)

6. Positive number before normalization:

0.000 164 779 749 346 824 7(10) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1001 0000 0011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 13 positions to the right, so that only one non zero digit remains to the left of it:


0.000 164 779 749 346 824 7(10) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1001 0000 0011 1(2) =

0.0000 0000 0000 1010 1100 1100 1000 1011 1010 0010 0110 1101 1100 1001 0000 0011 1(2) × 20 =

1.0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0010 0000 0111(2) × 2-13


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): -13

Mantissa (not normalized):
1.0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0010 0000 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-13 + 2(11-1) - 1 =

(-13 + 1 023)(10) =

1 010(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;
  • 1 010 ÷ 2 = 505 + 0;
  • 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) =

1010(10) =

011 1111 0010(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. 0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0010 0000 0111 =

0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0010 0000 0111


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 0010

Mantissa (52 bits) =
0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0010 0000 0111


Decimal number -0.000 164 779 749 346 824 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0010 - 0101 1001 1001 0001 0111 0100 0100 1101 1011 1001 0010 0000 0111

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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