-0.016 738 891 601 562 527 755 575 615 628 912 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 527 755 575 615 628 912 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
-0.016 738 891 601 562 527 755 575 615 628 912 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. Start with the positive version of the number:

|-0.016 738 891 601 562 527 755 575 615 628 912 9| = 0.016 738 891 601 562 527 755 575 615 628 912 9


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: 0.016 738 891 601 562 527 755 575 615 628 912 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) 0.016 738 891 601 562 527 755 575 615 628 912 9 × 2 = 0 + 0.033 477 783 203 125 055 511 151 231 257 825 8;
  • 2) 0.033 477 783 203 125 055 511 151 231 257 825 8 × 2 = 0 + 0.066 955 566 406 250 111 022 302 462 515 651 6;
  • 3) 0.066 955 566 406 250 111 022 302 462 515 651 6 × 2 = 0 + 0.133 911 132 812 500 222 044 604 925 031 303 2;
  • 4) 0.133 911 132 812 500 222 044 604 925 031 303 2 × 2 = 0 + 0.267 822 265 625 000 444 089 209 850 062 606 4;
  • 5) 0.267 822 265 625 000 444 089 209 850 062 606 4 × 2 = 0 + 0.535 644 531 250 000 888 178 419 700 125 212 8;
  • 6) 0.535 644 531 250 000 888 178 419 700 125 212 8 × 2 = 1 + 0.071 289 062 500 001 776 356 839 400 250 425 6;
  • 7) 0.071 289 062 500 001 776 356 839 400 250 425 6 × 2 = 0 + 0.142 578 125 000 003 552 713 678 800 500 851 2;
  • 8) 0.142 578 125 000 003 552 713 678 800 500 851 2 × 2 = 0 + 0.285 156 250 000 007 105 427 357 601 001 702 4;
  • 9) 0.285 156 250 000 007 105 427 357 601 001 702 4 × 2 = 0 + 0.570 312 500 000 014 210 854 715 202 003 404 8;
  • 10) 0.570 312 500 000 014 210 854 715 202 003 404 8 × 2 = 1 + 0.140 625 000 000 028 421 709 430 404 006 809 6;
  • 11) 0.140 625 000 000 028 421 709 430 404 006 809 6 × 2 = 0 + 0.281 250 000 000 056 843 418 860 808 013 619 2;
  • 12) 0.281 250 000 000 056 843 418 860 808 013 619 2 × 2 = 0 + 0.562 500 000 000 113 686 837 721 616 027 238 4;
  • 13) 0.562 500 000 000 113 686 837 721 616 027 238 4 × 2 = 1 + 0.125 000 000 000 227 373 675 443 232 054 476 8;
  • 14) 0.125 000 000 000 227 373 675 443 232 054 476 8 × 2 = 0 + 0.250 000 000 000 454 747 350 886 464 108 953 6;
  • 15) 0.250 000 000 000 454 747 350 886 464 108 953 6 × 2 = 0 + 0.500 000 000 000 909 494 701 772 928 217 907 2;
  • 16) 0.500 000 000 000 909 494 701 772 928 217 907 2 × 2 = 1 + 0.000 000 000 001 818 989 403 545 856 435 814 4;
  • 17) 0.000 000 000 001 818 989 403 545 856 435 814 4 × 2 = 0 + 0.000 000 000 003 637 978 807 091 712 871 628 8;
  • 18) 0.000 000 000 003 637 978 807 091 712 871 628 8 × 2 = 0 + 0.000 000 000 007 275 957 614 183 425 743 257 6;
  • 19) 0.000 000 000 007 275 957 614 183 425 743 257 6 × 2 = 0 + 0.000 000 000 014 551 915 228 366 851 486 515 2;
  • 20) 0.000 000 000 014 551 915 228 366 851 486 515 2 × 2 = 0 + 0.000 000 000 029 103 830 456 733 702 973 030 4;
  • 21) 0.000 000 000 029 103 830 456 733 702 973 030 4 × 2 = 0 + 0.000 000 000 058 207 660 913 467 405 946 060 8;
  • 22) 0.000 000 000 058 207 660 913 467 405 946 060 8 × 2 = 0 + 0.000 000 000 116 415 321 826 934 811 892 121 6;
  • 23) 0.000 000 000 116 415 321 826 934 811 892 121 6 × 2 = 0 + 0.000 000 000 232 830 643 653 869 623 784 243 2;
  • 24) 0.000 000 000 232 830 643 653 869 623 784 243 2 × 2 = 0 + 0.000 000 000 465 661 287 307 739 247 568 486 4;
  • 25) 0.000 000 000 465 661 287 307 739 247 568 486 4 × 2 = 0 + 0.000 000 000 931 322 574 615 478 495 136 972 8;
  • 26) 0.000 000 000 931 322 574 615 478 495 136 972 8 × 2 = 0 + 0.000 000 001 862 645 149 230 956 990 273 945 6;
  • 27) 0.000 000 001 862 645 149 230 956 990 273 945 6 × 2 = 0 + 0.000 000 003 725 290 298 461 913 980 547 891 2;
  • 28) 0.000 000 003 725 290 298 461 913 980 547 891 2 × 2 = 0 + 0.000 000 007 450 580 596 923 827 961 095 782 4;
  • 29) 0.000 000 007 450 580 596 923 827 961 095 782 4 × 2 = 0 + 0.000 000 014 901 161 193 847 655 922 191 564 8;
  • 30) 0.000 000 014 901 161 193 847 655 922 191 564 8 × 2 = 0 + 0.000 000 029 802 322 387 695 311 844 383 129 6;
  • 31) 0.000 000 029 802 322 387 695 311 844 383 129 6 × 2 = 0 + 0.000 000 059 604 644 775 390 623 688 766 259 2;
  • 32) 0.000 000 059 604 644 775 390 623 688 766 259 2 × 2 = 0 + 0.000 000 119 209 289 550 781 247 377 532 518 4;
  • 33) 0.000 000 119 209 289 550 781 247 377 532 518 4 × 2 = 0 + 0.000 000 238 418 579 101 562 494 755 065 036 8;
  • 34) 0.000 000 238 418 579 101 562 494 755 065 036 8 × 2 = 0 + 0.000 000 476 837 158 203 124 989 510 130 073 6;
  • 35) 0.000 000 476 837 158 203 124 989 510 130 073 6 × 2 = 0 + 0.000 000 953 674 316 406 249 979 020 260 147 2;
  • 36) 0.000 000 953 674 316 406 249 979 020 260 147 2 × 2 = 0 + 0.000 001 907 348 632 812 499 958 040 520 294 4;
  • 37) 0.000 001 907 348 632 812 499 958 040 520 294 4 × 2 = 0 + 0.000 003 814 697 265 624 999 916 081 040 588 8;
  • 38) 0.000 003 814 697 265 624 999 916 081 040 588 8 × 2 = 0 + 0.000 007 629 394 531 249 999 832 162 081 177 6;
  • 39) 0.000 007 629 394 531 249 999 832 162 081 177 6 × 2 = 0 + 0.000 015 258 789 062 499 999 664 324 162 355 2;
  • 40) 0.000 015 258 789 062 499 999 664 324 162 355 2 × 2 = 0 + 0.000 030 517 578 124 999 999 328 648 324 710 4;
  • 41) 0.000 030 517 578 124 999 999 328 648 324 710 4 × 2 = 0 + 0.000 061 035 156 249 999 998 657 296 649 420 8;
  • 42) 0.000 061 035 156 249 999 998 657 296 649 420 8 × 2 = 0 + 0.000 122 070 312 499 999 997 314 593 298 841 6;
  • 43) 0.000 122 070 312 499 999 997 314 593 298 841 6 × 2 = 0 + 0.000 244 140 624 999 999 994 629 186 597 683 2;
  • 44) 0.000 244 140 624 999 999 994 629 186 597 683 2 × 2 = 0 + 0.000 488 281 249 999 999 989 258 373 195 366 4;
  • 45) 0.000 488 281 249 999 999 989 258 373 195 366 4 × 2 = 0 + 0.000 976 562 499 999 999 978 516 746 390 732 8;
  • 46) 0.000 976 562 499 999 999 978 516 746 390 732 8 × 2 = 0 + 0.001 953 124 999 999 999 957 033 492 781 465 6;
  • 47) 0.001 953 124 999 999 999 957 033 492 781 465 6 × 2 = 0 + 0.003 906 249 999 999 999 914 066 985 562 931 2;
  • 48) 0.003 906 249 999 999 999 914 066 985 562 931 2 × 2 = 0 + 0.007 812 499 999 999 999 828 133 971 125 862 4;
  • 49) 0.007 812 499 999 999 999 828 133 971 125 862 4 × 2 = 0 + 0.015 624 999 999 999 999 656 267 942 251 724 8;
  • 50) 0.015 624 999 999 999 999 656 267 942 251 724 8 × 2 = 0 + 0.031 249 999 999 999 999 312 535 884 503 449 6;
  • 51) 0.031 249 999 999 999 999 312 535 884 503 449 6 × 2 = 0 + 0.062 499 999 999 999 998 625 071 769 006 899 2;
  • 52) 0.062 499 999 999 999 998 625 071 769 006 899 2 × 2 = 0 + 0.124 999 999 999 999 997 250 143 538 013 798 4;
  • 53) 0.124 999 999 999 999 997 250 143 538 013 798 4 × 2 = 0 + 0.249 999 999 999 999 994 500 287 076 027 596 8;
  • 54) 0.249 999 999 999 999 994 500 287 076 027 596 8 × 2 = 0 + 0.499 999 999 999 999 989 000 574 152 055 193 6;
  • 55) 0.499 999 999 999 999 989 000 574 152 055 193 6 × 2 = 0 + 0.999 999 999 999 999 978 001 148 304 110 387 2;
  • 56) 0.999 999 999 999 999 978 001 148 304 110 387 2 × 2 = 1 + 0.999 999 999 999 999 956 002 296 608 220 774 4;
  • 57) 0.999 999 999 999 999 956 002 296 608 220 774 4 × 2 = 1 + 0.999 999 999 999 999 912 004 593 216 441 548 8;
  • 58) 0.999 999 999 999 999 912 004 593 216 441 548 8 × 2 = 1 + 0.999 999 999 999 999 824 009 186 432 883 097 6;

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:


0.016 738 891 601 562 527 755 575 615 628 912 9(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 527 755 575 615 628 912 9(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 527 755 575 615 628 912 9(10) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 11(2) =

0.0000 0100 0100 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 11(2) × 20 =

1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1017(10) =

011 1111 1001(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. 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 =

0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 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 1001

Mantissa (52 bits) =
0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111


Decimal number -0.016 738 891 601 562 527 755 575 615 628 912 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0100 0000 0000 0000 0000 0000 0000 0000 0000 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