0.000 000 000 000 000 444 092 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 000 444 092(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
0.000 000 000 000 000 444 092(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: 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 444 092.

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) 0.000 000 000 000 000 444 092 × 2 = 0 + 0.000 000 000 000 000 888 184;
  • 2) 0.000 000 000 000 000 888 184 × 2 = 0 + 0.000 000 000 000 001 776 368;
  • 3) 0.000 000 000 000 001 776 368 × 2 = 0 + 0.000 000 000 000 003 552 736;
  • 4) 0.000 000 000 000 003 552 736 × 2 = 0 + 0.000 000 000 000 007 105 472;
  • 5) 0.000 000 000 000 007 105 472 × 2 = 0 + 0.000 000 000 000 014 210 944;
  • 6) 0.000 000 000 000 014 210 944 × 2 = 0 + 0.000 000 000 000 028 421 888;
  • 7) 0.000 000 000 000 028 421 888 × 2 = 0 + 0.000 000 000 000 056 843 776;
  • 8) 0.000 000 000 000 056 843 776 × 2 = 0 + 0.000 000 000 000 113 687 552;
  • 9) 0.000 000 000 000 113 687 552 × 2 = 0 + 0.000 000 000 000 227 375 104;
  • 10) 0.000 000 000 000 227 375 104 × 2 = 0 + 0.000 000 000 000 454 750 208;
  • 11) 0.000 000 000 000 454 750 208 × 2 = 0 + 0.000 000 000 000 909 500 416;
  • 12) 0.000 000 000 000 909 500 416 × 2 = 0 + 0.000 000 000 001 819 000 832;
  • 13) 0.000 000 000 001 819 000 832 × 2 = 0 + 0.000 000 000 003 638 001 664;
  • 14) 0.000 000 000 003 638 001 664 × 2 = 0 + 0.000 000 000 007 276 003 328;
  • 15) 0.000 000 000 007 276 003 328 × 2 = 0 + 0.000 000 000 014 552 006 656;
  • 16) 0.000 000 000 014 552 006 656 × 2 = 0 + 0.000 000 000 029 104 013 312;
  • 17) 0.000 000 000 029 104 013 312 × 2 = 0 + 0.000 000 000 058 208 026 624;
  • 18) 0.000 000 000 058 208 026 624 × 2 = 0 + 0.000 000 000 116 416 053 248;
  • 19) 0.000 000 000 116 416 053 248 × 2 = 0 + 0.000 000 000 232 832 106 496;
  • 20) 0.000 000 000 232 832 106 496 × 2 = 0 + 0.000 000 000 465 664 212 992;
  • 21) 0.000 000 000 465 664 212 992 × 2 = 0 + 0.000 000 000 931 328 425 984;
  • 22) 0.000 000 000 931 328 425 984 × 2 = 0 + 0.000 000 001 862 656 851 968;
  • 23) 0.000 000 001 862 656 851 968 × 2 = 0 + 0.000 000 003 725 313 703 936;
  • 24) 0.000 000 003 725 313 703 936 × 2 = 0 + 0.000 000 007 450 627 407 872;
  • 25) 0.000 000 007 450 627 407 872 × 2 = 0 + 0.000 000 014 901 254 815 744;
  • 26) 0.000 000 014 901 254 815 744 × 2 = 0 + 0.000 000 029 802 509 631 488;
  • 27) 0.000 000 029 802 509 631 488 × 2 = 0 + 0.000 000 059 605 019 262 976;
  • 28) 0.000 000 059 605 019 262 976 × 2 = 0 + 0.000 000 119 210 038 525 952;
  • 29) 0.000 000 119 210 038 525 952 × 2 = 0 + 0.000 000 238 420 077 051 904;
  • 30) 0.000 000 238 420 077 051 904 × 2 = 0 + 0.000 000 476 840 154 103 808;
  • 31) 0.000 000 476 840 154 103 808 × 2 = 0 + 0.000 000 953 680 308 207 616;
  • 32) 0.000 000 953 680 308 207 616 × 2 = 0 + 0.000 001 907 360 616 415 232;
  • 33) 0.000 001 907 360 616 415 232 × 2 = 0 + 0.000 003 814 721 232 830 464;
  • 34) 0.000 003 814 721 232 830 464 × 2 = 0 + 0.000 007 629 442 465 660 928;
  • 35) 0.000 007 629 442 465 660 928 × 2 = 0 + 0.000 015 258 884 931 321 856;
  • 36) 0.000 015 258 884 931 321 856 × 2 = 0 + 0.000 030 517 769 862 643 712;
  • 37) 0.000 030 517 769 862 643 712 × 2 = 0 + 0.000 061 035 539 725 287 424;
  • 38) 0.000 061 035 539 725 287 424 × 2 = 0 + 0.000 122 071 079 450 574 848;
  • 39) 0.000 122 071 079 450 574 848 × 2 = 0 + 0.000 244 142 158 901 149 696;
  • 40) 0.000 244 142 158 901 149 696 × 2 = 0 + 0.000 488 284 317 802 299 392;
  • 41) 0.000 488 284 317 802 299 392 × 2 = 0 + 0.000 976 568 635 604 598 784;
  • 42) 0.000 976 568 635 604 598 784 × 2 = 0 + 0.001 953 137 271 209 197 568;
  • 43) 0.001 953 137 271 209 197 568 × 2 = 0 + 0.003 906 274 542 418 395 136;
  • 44) 0.003 906 274 542 418 395 136 × 2 = 0 + 0.007 812 549 084 836 790 272;
  • 45) 0.007 812 549 084 836 790 272 × 2 = 0 + 0.015 625 098 169 673 580 544;
  • 46) 0.015 625 098 169 673 580 544 × 2 = 0 + 0.031 250 196 339 347 161 088;
  • 47) 0.031 250 196 339 347 161 088 × 2 = 0 + 0.062 500 392 678 694 322 176;
  • 48) 0.062 500 392 678 694 322 176 × 2 = 0 + 0.125 000 785 357 388 644 352;
  • 49) 0.125 000 785 357 388 644 352 × 2 = 0 + 0.250 001 570 714 777 288 704;
  • 50) 0.250 001 570 714 777 288 704 × 2 = 0 + 0.500 003 141 429 554 577 408;
  • 51) 0.500 003 141 429 554 577 408 × 2 = 1 + 0.000 006 282 859 109 154 816;
  • 52) 0.000 006 282 859 109 154 816 × 2 = 0 + 0.000 012 565 718 218 309 632;
  • 53) 0.000 012 565 718 218 309 632 × 2 = 0 + 0.000 025 131 436 436 619 264;
  • 54) 0.000 025 131 436 436 619 264 × 2 = 0 + 0.000 050 262 872 873 238 528;
  • 55) 0.000 050 262 872 873 238 528 × 2 = 0 + 0.000 100 525 745 746 477 056;
  • 56) 0.000 100 525 745 746 477 056 × 2 = 0 + 0.000 201 051 491 492 954 112;
  • 57) 0.000 201 051 491 492 954 112 × 2 = 0 + 0.000 402 102 982 985 908 224;
  • 58) 0.000 402 102 982 985 908 224 × 2 = 0 + 0.000 804 205 965 971 816 448;
  • 59) 0.000 804 205 965 971 816 448 × 2 = 0 + 0.001 608 411 931 943 632 896;
  • 60) 0.001 608 411 931 943 632 896 × 2 = 0 + 0.003 216 823 863 887 265 792;
  • 61) 0.003 216 823 863 887 265 792 × 2 = 0 + 0.006 433 647 727 774 531 584;
  • 62) 0.006 433 647 727 774 531 584 × 2 = 0 + 0.012 867 295 455 549 063 168;
  • 63) 0.012 867 295 455 549 063 168 × 2 = 0 + 0.025 734 590 911 098 126 336;
  • 64) 0.025 734 590 911 098 126 336 × 2 = 0 + 0.051 469 181 822 196 252 672;
  • 65) 0.051 469 181 822 196 252 672 × 2 = 0 + 0.102 938 363 644 392 505 344;
  • 66) 0.102 938 363 644 392 505 344 × 2 = 0 + 0.205 876 727 288 785 010 688;
  • 67) 0.205 876 727 288 785 010 688 × 2 = 0 + 0.411 753 454 577 570 021 376;
  • 68) 0.411 753 454 577 570 021 376 × 2 = 0 + 0.823 506 909 155 140 042 752;
  • 69) 0.823 506 909 155 140 042 752 × 2 = 1 + 0.647 013 818 310 280 085 504;
  • 70) 0.647 013 818 310 280 085 504 × 2 = 1 + 0.294 027 636 620 560 171 008;
  • 71) 0.294 027 636 620 560 171 008 × 2 = 0 + 0.588 055 273 241 120 342 016;
  • 72) 0.588 055 273 241 120 342 016 × 2 = 1 + 0.176 110 546 482 240 684 032;
  • 73) 0.176 110 546 482 240 684 032 × 2 = 0 + 0.352 221 092 964 481 368 064;
  • 74) 0.352 221 092 964 481 368 064 × 2 = 0 + 0.704 442 185 928 962 736 128;
  • 75) 0.704 442 185 928 962 736 128 × 2 = 1 + 0.408 884 371 857 925 472 256;
  • 76) 0.408 884 371 857 925 472 256 × 2 = 0 + 0.817 768 743 715 850 944 512;
  • 77) 0.817 768 743 715 850 944 512 × 2 = 1 + 0.635 537 487 431 701 889 024;
  • 78) 0.635 537 487 431 701 889 024 × 2 = 1 + 0.271 074 974 863 403 778 048;
  • 79) 0.271 074 974 863 403 778 048 × 2 = 0 + 0.542 149 949 726 807 556 096;
  • 80) 0.542 149 949 726 807 556 096 × 2 = 1 + 0.084 299 899 453 615 112 192;
  • 81) 0.084 299 899 453 615 112 192 × 2 = 0 + 0.168 599 798 907 230 224 384;
  • 82) 0.168 599 798 907 230 224 384 × 2 = 0 + 0.337 199 597 814 460 448 768;
  • 83) 0.337 199 597 814 460 448 768 × 2 = 0 + 0.674 399 195 628 920 897 536;
  • 84) 0.674 399 195 628 920 897 536 × 2 = 1 + 0.348 798 391 257 841 795 072;
  • 85) 0.348 798 391 257 841 795 072 × 2 = 0 + 0.697 596 782 515 683 590 144;
  • 86) 0.697 596 782 515 683 590 144 × 2 = 1 + 0.395 193 565 031 367 180 288;
  • 87) 0.395 193 565 031 367 180 288 × 2 = 0 + 0.790 387 130 062 734 360 576;
  • 88) 0.790 387 130 062 734 360 576 × 2 = 1 + 0.580 774 260 125 468 721 152;
  • 89) 0.580 774 260 125 468 721 152 × 2 = 1 + 0.161 548 520 250 937 442 304;
  • 90) 0.161 548 520 250 937 442 304 × 2 = 0 + 0.323 097 040 501 874 884 608;
  • 91) 0.323 097 040 501 874 884 608 × 2 = 0 + 0.646 194 081 003 749 769 216;
  • 92) 0.646 194 081 003 749 769 216 × 2 = 1 + 0.292 388 162 007 499 538 432;
  • 93) 0.292 388 162 007 499 538 432 × 2 = 0 + 0.584 776 324 014 999 076 864;
  • 94) 0.584 776 324 014 999 076 864 × 2 = 1 + 0.169 552 648 029 998 153 728;
  • 95) 0.169 552 648 029 998 153 728 × 2 = 0 + 0.339 105 296 059 996 307 456;
  • 96) 0.339 105 296 059 996 307 456 × 2 = 0 + 0.678 210 592 119 992 614 912;
  • 97) 0.678 210 592 119 992 614 912 × 2 = 1 + 0.356 421 184 239 985 229 824;
  • 98) 0.356 421 184 239 985 229 824 × 2 = 0 + 0.712 842 368 479 970 459 648;
  • 99) 0.712 842 368 479 970 459 648 × 2 = 1 + 0.425 684 736 959 940 919 296;
  • 100) 0.425 684 736 959 940 919 296 × 2 = 0 + 0.851 369 473 919 881 838 592;
  • 101) 0.851 369 473 919 881 838 592 × 2 = 1 + 0.702 738 947 839 763 677 184;
  • 102) 0.702 738 947 839 763 677 184 × 2 = 1 + 0.405 477 895 679 527 354 368;
  • 103) 0.405 477 895 679 527 354 368 × 2 = 0 + 0.810 955 791 359 054 708 736;

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:


0.000 000 000 000 000 444 092(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0000 0000 0000 1101 0010 1101 0001 0101 1001 0100 1010 110(2)

5. Positive number before normalization:

0.000 000 000 000 000 444 092(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0000 0000 0000 1101 0010 1101 0001 0101 1001 0100 1010 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 51 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 444 092(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0000 0000 0000 1101 0010 1101 0001 0101 1001 0100 1010 110(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0000 0000 0000 1101 0010 1101 0001 0101 1001 0100 1010 110(2) × 20 =

1.0000 0000 0000 0000 0110 1001 0110 1000 1010 1100 1010 0101 0110(2) × 2-51


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): -51

Mantissa (not normalized):
1.0000 0000 0000 0000 0110 1001 0110 1000 1010 1100 1010 0101 0110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-51 + 2(11-1) - 1 =

(-51 + 1 023)(10) =

972(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;
  • 972 ÷ 2 = 486 + 0;
  • 486 ÷ 2 = 243 + 0;
  • 243 ÷ 2 = 121 + 1;
  • 121 ÷ 2 = 60 + 1;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

972(10) =

011 1100 1100(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0000 0000 0000 0000 0110 1001 0110 1000 1010 1100 1010 0101 0110 =

0000 0000 0000 0000 0110 1001 0110 1000 1010 1100 1010 0101 0110


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) =
011 1100 1100

Mantissa (52 bits) =
0000 0000 0000 0000 0110 1001 0110 1000 1010 1100 1010 0101 0110


Decimal number 0.000 000 000 000 000 444 092 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1100 1100 - 0000 0000 0000 0000 0110 1001 0110 1000 1010 1100 1010 0101 0110

» Convert decimal 0.000 000 000 000 000 444 059 to 64 bit double precision IEEE 754 binary floating point representation standard

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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