-0.000 000 017 188 905 742 259 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 017 188 905 742 259 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
-0.000 000 017 188 905 742 259 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:

|-0.000 000 017 188 905 742 259 53| = 0.000 000 017 188 905 742 259 53


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: 0.000 000 017 188 905 742 259 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) 0.000 000 017 188 905 742 259 53 × 2 = 0 + 0.000 000 034 377 811 484 519 06;
  • 2) 0.000 000 034 377 811 484 519 06 × 2 = 0 + 0.000 000 068 755 622 969 038 12;
  • 3) 0.000 000 068 755 622 969 038 12 × 2 = 0 + 0.000 000 137 511 245 938 076 24;
  • 4) 0.000 000 137 511 245 938 076 24 × 2 = 0 + 0.000 000 275 022 491 876 152 48;
  • 5) 0.000 000 275 022 491 876 152 48 × 2 = 0 + 0.000 000 550 044 983 752 304 96;
  • 6) 0.000 000 550 044 983 752 304 96 × 2 = 0 + 0.000 001 100 089 967 504 609 92;
  • 7) 0.000 001 100 089 967 504 609 92 × 2 = 0 + 0.000 002 200 179 935 009 219 84;
  • 8) 0.000 002 200 179 935 009 219 84 × 2 = 0 + 0.000 004 400 359 870 018 439 68;
  • 9) 0.000 004 400 359 870 018 439 68 × 2 = 0 + 0.000 008 800 719 740 036 879 36;
  • 10) 0.000 008 800 719 740 036 879 36 × 2 = 0 + 0.000 017 601 439 480 073 758 72;
  • 11) 0.000 017 601 439 480 073 758 72 × 2 = 0 + 0.000 035 202 878 960 147 517 44;
  • 12) 0.000 035 202 878 960 147 517 44 × 2 = 0 + 0.000 070 405 757 920 295 034 88;
  • 13) 0.000 070 405 757 920 295 034 88 × 2 = 0 + 0.000 140 811 515 840 590 069 76;
  • 14) 0.000 140 811 515 840 590 069 76 × 2 = 0 + 0.000 281 623 031 681 180 139 52;
  • 15) 0.000 281 623 031 681 180 139 52 × 2 = 0 + 0.000 563 246 063 362 360 279 04;
  • 16) 0.000 563 246 063 362 360 279 04 × 2 = 0 + 0.001 126 492 126 724 720 558 08;
  • 17) 0.001 126 492 126 724 720 558 08 × 2 = 0 + 0.002 252 984 253 449 441 116 16;
  • 18) 0.002 252 984 253 449 441 116 16 × 2 = 0 + 0.004 505 968 506 898 882 232 32;
  • 19) 0.004 505 968 506 898 882 232 32 × 2 = 0 + 0.009 011 937 013 797 764 464 64;
  • 20) 0.009 011 937 013 797 764 464 64 × 2 = 0 + 0.018 023 874 027 595 528 929 28;
  • 21) 0.018 023 874 027 595 528 929 28 × 2 = 0 + 0.036 047 748 055 191 057 858 56;
  • 22) 0.036 047 748 055 191 057 858 56 × 2 = 0 + 0.072 095 496 110 382 115 717 12;
  • 23) 0.072 095 496 110 382 115 717 12 × 2 = 0 + 0.144 190 992 220 764 231 434 24;
  • 24) 0.144 190 992 220 764 231 434 24 × 2 = 0 + 0.288 381 984 441 528 462 868 48;
  • 25) 0.288 381 984 441 528 462 868 48 × 2 = 0 + 0.576 763 968 883 056 925 736 96;
  • 26) 0.576 763 968 883 056 925 736 96 × 2 = 1 + 0.153 527 937 766 113 851 473 92;
  • 27) 0.153 527 937 766 113 851 473 92 × 2 = 0 + 0.307 055 875 532 227 702 947 84;
  • 28) 0.307 055 875 532 227 702 947 84 × 2 = 0 + 0.614 111 751 064 455 405 895 68;
  • 29) 0.614 111 751 064 455 405 895 68 × 2 = 1 + 0.228 223 502 128 910 811 791 36;
  • 30) 0.228 223 502 128 910 811 791 36 × 2 = 0 + 0.456 447 004 257 821 623 582 72;
  • 31) 0.456 447 004 257 821 623 582 72 × 2 = 0 + 0.912 894 008 515 643 247 165 44;
  • 32) 0.912 894 008 515 643 247 165 44 × 2 = 1 + 0.825 788 017 031 286 494 330 88;
  • 33) 0.825 788 017 031 286 494 330 88 × 2 = 1 + 0.651 576 034 062 572 988 661 76;
  • 34) 0.651 576 034 062 572 988 661 76 × 2 = 1 + 0.303 152 068 125 145 977 323 52;
  • 35) 0.303 152 068 125 145 977 323 52 × 2 = 0 + 0.606 304 136 250 291 954 647 04;
  • 36) 0.606 304 136 250 291 954 647 04 × 2 = 1 + 0.212 608 272 500 583 909 294 08;
  • 37) 0.212 608 272 500 583 909 294 08 × 2 = 0 + 0.425 216 545 001 167 818 588 16;
  • 38) 0.425 216 545 001 167 818 588 16 × 2 = 0 + 0.850 433 090 002 335 637 176 32;
  • 39) 0.850 433 090 002 335 637 176 32 × 2 = 1 + 0.700 866 180 004 671 274 352 64;
  • 40) 0.700 866 180 004 671 274 352 64 × 2 = 1 + 0.401 732 360 009 342 548 705 28;
  • 41) 0.401 732 360 009 342 548 705 28 × 2 = 0 + 0.803 464 720 018 685 097 410 56;
  • 42) 0.803 464 720 018 685 097 410 56 × 2 = 1 + 0.606 929 440 037 370 194 821 12;
  • 43) 0.606 929 440 037 370 194 821 12 × 2 = 1 + 0.213 858 880 074 740 389 642 24;
  • 44) 0.213 858 880 074 740 389 642 24 × 2 = 0 + 0.427 717 760 149 480 779 284 48;
  • 45) 0.427 717 760 149 480 779 284 48 × 2 = 0 + 0.855 435 520 298 961 558 568 96;
  • 46) 0.855 435 520 298 961 558 568 96 × 2 = 1 + 0.710 871 040 597 923 117 137 92;
  • 47) 0.710 871 040 597 923 117 137 92 × 2 = 1 + 0.421 742 081 195 846 234 275 84;
  • 48) 0.421 742 081 195 846 234 275 84 × 2 = 0 + 0.843 484 162 391 692 468 551 68;
  • 49) 0.843 484 162 391 692 468 551 68 × 2 = 1 + 0.686 968 324 783 384 937 103 36;
  • 50) 0.686 968 324 783 384 937 103 36 × 2 = 1 + 0.373 936 649 566 769 874 206 72;
  • 51) 0.373 936 649 566 769 874 206 72 × 2 = 0 + 0.747 873 299 133 539 748 413 44;
  • 52) 0.747 873 299 133 539 748 413 44 × 2 = 1 + 0.495 746 598 267 079 496 826 88;
  • 53) 0.495 746 598 267 079 496 826 88 × 2 = 0 + 0.991 493 196 534 158 993 653 76;
  • 54) 0.991 493 196 534 158 993 653 76 × 2 = 1 + 0.982 986 393 068 317 987 307 52;
  • 55) 0.982 986 393 068 317 987 307 52 × 2 = 1 + 0.965 972 786 136 635 974 615 04;
  • 56) 0.965 972 786 136 635 974 615 04 × 2 = 1 + 0.931 945 572 273 271 949 230 08;
  • 57) 0.931 945 572 273 271 949 230 08 × 2 = 1 + 0.863 891 144 546 543 898 460 16;
  • 58) 0.863 891 144 546 543 898 460 16 × 2 = 1 + 0.727 782 289 093 087 796 920 32;
  • 59) 0.727 782 289 093 087 796 920 32 × 2 = 1 + 0.455 564 578 186 175 593 840 64;
  • 60) 0.455 564 578 186 175 593 840 64 × 2 = 0 + 0.911 129 156 372 351 187 681 28;
  • 61) 0.911 129 156 372 351 187 681 28 × 2 = 1 + 0.822 258 312 744 702 375 362 56;
  • 62) 0.822 258 312 744 702 375 362 56 × 2 = 1 + 0.644 516 625 489 404 750 725 12;
  • 63) 0.644 516 625 489 404 750 725 12 × 2 = 1 + 0.289 033 250 978 809 501 450 24;
  • 64) 0.289 033 250 978 809 501 450 24 × 2 = 0 + 0.578 066 501 957 619 002 900 48;
  • 65) 0.578 066 501 957 619 002 900 48 × 2 = 1 + 0.156 133 003 915 238 005 800 96;
  • 66) 0.156 133 003 915 238 005 800 96 × 2 = 0 + 0.312 266 007 830 476 011 601 92;
  • 67) 0.312 266 007 830 476 011 601 92 × 2 = 0 + 0.624 532 015 660 952 023 203 84;
  • 68) 0.624 532 015 660 952 023 203 84 × 2 = 1 + 0.249 064 031 321 904 046 407 68;
  • 69) 0.249 064 031 321 904 046 407 68 × 2 = 0 + 0.498 128 062 643 808 092 815 36;
  • 70) 0.498 128 062 643 808 092 815 36 × 2 = 0 + 0.996 256 125 287 616 185 630 72;
  • 71) 0.996 256 125 287 616 185 630 72 × 2 = 1 + 0.992 512 250 575 232 371 261 44;
  • 72) 0.992 512 250 575 232 371 261 44 × 2 = 1 + 0.985 024 501 150 464 742 522 88;
  • 73) 0.985 024 501 150 464 742 522 88 × 2 = 1 + 0.970 049 002 300 929 485 045 76;
  • 74) 0.970 049 002 300 929 485 045 76 × 2 = 1 + 0.940 098 004 601 858 970 091 52;
  • 75) 0.940 098 004 601 858 970 091 52 × 2 = 1 + 0.880 196 009 203 717 940 183 04;
  • 76) 0.880 196 009 203 717 940 183 04 × 2 = 1 + 0.760 392 018 407 435 880 366 08;
  • 77) 0.760 392 018 407 435 880 366 08 × 2 = 1 + 0.520 784 036 814 871 760 732 16;
  • 78) 0.520 784 036 814 871 760 732 16 × 2 = 1 + 0.041 568 073 629 743 521 464 32;

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:


0.000 000 017 188 905 742 259 53(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0011 1111 11(2)

6. Positive number before normalization:

0.000 000 017 188 905 742 259 53(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0011 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 017 188 905 742 259 53(10) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0011 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0100 1001 1101 0011 0110 0110 1101 0111 1110 1110 1001 0011 1111 11(2) × 20 =

1.0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 1111 1111(2) × 2-26


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): -26

Mantissa (not normalized):
1.0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(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;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

997(10) =

011 1110 0101(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. 0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 1111 1111 =

0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 1111 1111


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 1110 0101

Mantissa (52 bits) =
0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 1111 1111


Decimal number -0.000 000 017 188 905 742 259 53 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0101 - 0010 0111 0100 1101 1001 1011 0101 1111 1011 1010 0100 1111 1111

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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