-0.000 000 045 419 381 005 632 303 53 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 045 419 381 005 632 303 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
-0.000 000 045 419 381 005 632 303 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:

|-0.000 000 045 419 381 005 632 303 53| = 0.000 000 045 419 381 005 632 303 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: 0.000 000 045 419 381 005 632 303 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) 0.000 000 045 419 381 005 632 303 53 × 2 = 0 + 0.000 000 090 838 762 011 264 607 06;
  • 2) 0.000 000 090 838 762 011 264 607 06 × 2 = 0 + 0.000 000 181 677 524 022 529 214 12;
  • 3) 0.000 000 181 677 524 022 529 214 12 × 2 = 0 + 0.000 000 363 355 048 045 058 428 24;
  • 4) 0.000 000 363 355 048 045 058 428 24 × 2 = 0 + 0.000 000 726 710 096 090 116 856 48;
  • 5) 0.000 000 726 710 096 090 116 856 48 × 2 = 0 + 0.000 001 453 420 192 180 233 712 96;
  • 6) 0.000 001 453 420 192 180 233 712 96 × 2 = 0 + 0.000 002 906 840 384 360 467 425 92;
  • 7) 0.000 002 906 840 384 360 467 425 92 × 2 = 0 + 0.000 005 813 680 768 720 934 851 84;
  • 8) 0.000 005 813 680 768 720 934 851 84 × 2 = 0 + 0.000 011 627 361 537 441 869 703 68;
  • 9) 0.000 011 627 361 537 441 869 703 68 × 2 = 0 + 0.000 023 254 723 074 883 739 407 36;
  • 10) 0.000 023 254 723 074 883 739 407 36 × 2 = 0 + 0.000 046 509 446 149 767 478 814 72;
  • 11) 0.000 046 509 446 149 767 478 814 72 × 2 = 0 + 0.000 093 018 892 299 534 957 629 44;
  • 12) 0.000 093 018 892 299 534 957 629 44 × 2 = 0 + 0.000 186 037 784 599 069 915 258 88;
  • 13) 0.000 186 037 784 599 069 915 258 88 × 2 = 0 + 0.000 372 075 569 198 139 830 517 76;
  • 14) 0.000 372 075 569 198 139 830 517 76 × 2 = 0 + 0.000 744 151 138 396 279 661 035 52;
  • 15) 0.000 744 151 138 396 279 661 035 52 × 2 = 0 + 0.001 488 302 276 792 559 322 071 04;
  • 16) 0.001 488 302 276 792 559 322 071 04 × 2 = 0 + 0.002 976 604 553 585 118 644 142 08;
  • 17) 0.002 976 604 553 585 118 644 142 08 × 2 = 0 + 0.005 953 209 107 170 237 288 284 16;
  • 18) 0.005 953 209 107 170 237 288 284 16 × 2 = 0 + 0.011 906 418 214 340 474 576 568 32;
  • 19) 0.011 906 418 214 340 474 576 568 32 × 2 = 0 + 0.023 812 836 428 680 949 153 136 64;
  • 20) 0.023 812 836 428 680 949 153 136 64 × 2 = 0 + 0.047 625 672 857 361 898 306 273 28;
  • 21) 0.047 625 672 857 361 898 306 273 28 × 2 = 0 + 0.095 251 345 714 723 796 612 546 56;
  • 22) 0.095 251 345 714 723 796 612 546 56 × 2 = 0 + 0.190 502 691 429 447 593 225 093 12;
  • 23) 0.190 502 691 429 447 593 225 093 12 × 2 = 0 + 0.381 005 382 858 895 186 450 186 24;
  • 24) 0.381 005 382 858 895 186 450 186 24 × 2 = 0 + 0.762 010 765 717 790 372 900 372 48;
  • 25) 0.762 010 765 717 790 372 900 372 48 × 2 = 1 + 0.524 021 531 435 580 745 800 744 96;
  • 26) 0.524 021 531 435 580 745 800 744 96 × 2 = 1 + 0.048 043 062 871 161 491 601 489 92;
  • 27) 0.048 043 062 871 161 491 601 489 92 × 2 = 0 + 0.096 086 125 742 322 983 202 979 84;
  • 28) 0.096 086 125 742 322 983 202 979 84 × 2 = 0 + 0.192 172 251 484 645 966 405 959 68;
  • 29) 0.192 172 251 484 645 966 405 959 68 × 2 = 0 + 0.384 344 502 969 291 932 811 919 36;
  • 30) 0.384 344 502 969 291 932 811 919 36 × 2 = 0 + 0.768 689 005 938 583 865 623 838 72;
  • 31) 0.768 689 005 938 583 865 623 838 72 × 2 = 1 + 0.537 378 011 877 167 731 247 677 44;
  • 32) 0.537 378 011 877 167 731 247 677 44 × 2 = 1 + 0.074 756 023 754 335 462 495 354 88;
  • 33) 0.074 756 023 754 335 462 495 354 88 × 2 = 0 + 0.149 512 047 508 670 924 990 709 76;
  • 34) 0.149 512 047 508 670 924 990 709 76 × 2 = 0 + 0.299 024 095 017 341 849 981 419 52;
  • 35) 0.299 024 095 017 341 849 981 419 52 × 2 = 0 + 0.598 048 190 034 683 699 962 839 04;
  • 36) 0.598 048 190 034 683 699 962 839 04 × 2 = 1 + 0.196 096 380 069 367 399 925 678 08;
  • 37) 0.196 096 380 069 367 399 925 678 08 × 2 = 0 + 0.392 192 760 138 734 799 851 356 16;
  • 38) 0.392 192 760 138 734 799 851 356 16 × 2 = 0 + 0.784 385 520 277 469 599 702 712 32;
  • 39) 0.784 385 520 277 469 599 702 712 32 × 2 = 1 + 0.568 771 040 554 939 199 405 424 64;
  • 40) 0.568 771 040 554 939 199 405 424 64 × 2 = 1 + 0.137 542 081 109 878 398 810 849 28;
  • 41) 0.137 542 081 109 878 398 810 849 28 × 2 = 0 + 0.275 084 162 219 756 797 621 698 56;
  • 42) 0.275 084 162 219 756 797 621 698 56 × 2 = 0 + 0.550 168 324 439 513 595 243 397 12;
  • 43) 0.550 168 324 439 513 595 243 397 12 × 2 = 1 + 0.100 336 648 879 027 190 486 794 24;
  • 44) 0.100 336 648 879 027 190 486 794 24 × 2 = 0 + 0.200 673 297 758 054 380 973 588 48;
  • 45) 0.200 673 297 758 054 380 973 588 48 × 2 = 0 + 0.401 346 595 516 108 761 947 176 96;
  • 46) 0.401 346 595 516 108 761 947 176 96 × 2 = 0 + 0.802 693 191 032 217 523 894 353 92;
  • 47) 0.802 693 191 032 217 523 894 353 92 × 2 = 1 + 0.605 386 382 064 435 047 788 707 84;
  • 48) 0.605 386 382 064 435 047 788 707 84 × 2 = 1 + 0.210 772 764 128 870 095 577 415 68;
  • 49) 0.210 772 764 128 870 095 577 415 68 × 2 = 0 + 0.421 545 528 257 740 191 154 831 36;
  • 50) 0.421 545 528 257 740 191 154 831 36 × 2 = 0 + 0.843 091 056 515 480 382 309 662 72;
  • 51) 0.843 091 056 515 480 382 309 662 72 × 2 = 1 + 0.686 182 113 030 960 764 619 325 44;
  • 52) 0.686 182 113 030 960 764 619 325 44 × 2 = 1 + 0.372 364 226 061 921 529 238 650 88;
  • 53) 0.372 364 226 061 921 529 238 650 88 × 2 = 0 + 0.744 728 452 123 843 058 477 301 76;
  • 54) 0.744 728 452 123 843 058 477 301 76 × 2 = 1 + 0.489 456 904 247 686 116 954 603 52;
  • 55) 0.489 456 904 247 686 116 954 603 52 × 2 = 0 + 0.978 913 808 495 372 233 909 207 04;
  • 56) 0.978 913 808 495 372 233 909 207 04 × 2 = 1 + 0.957 827 616 990 744 467 818 414 08;
  • 57) 0.957 827 616 990 744 467 818 414 08 × 2 = 1 + 0.915 655 233 981 488 935 636 828 16;
  • 58) 0.915 655 233 981 488 935 636 828 16 × 2 = 1 + 0.831 310 467 962 977 871 273 656 32;
  • 59) 0.831 310 467 962 977 871 273 656 32 × 2 = 1 + 0.662 620 935 925 955 742 547 312 64;
  • 60) 0.662 620 935 925 955 742 547 312 64 × 2 = 1 + 0.325 241 871 851 911 485 094 625 28;
  • 61) 0.325 241 871 851 911 485 094 625 28 × 2 = 0 + 0.650 483 743 703 822 970 189 250 56;
  • 62) 0.650 483 743 703 822 970 189 250 56 × 2 = 1 + 0.300 967 487 407 645 940 378 501 12;
  • 63) 0.300 967 487 407 645 940 378 501 12 × 2 = 0 + 0.601 934 974 815 291 880 757 002 24;
  • 64) 0.601 934 974 815 291 880 757 002 24 × 2 = 1 + 0.203 869 949 630 583 761 514 004 48;
  • 65) 0.203 869 949 630 583 761 514 004 48 × 2 = 0 + 0.407 739 899 261 167 523 028 008 96;
  • 66) 0.407 739 899 261 167 523 028 008 96 × 2 = 0 + 0.815 479 798 522 335 046 056 017 92;
  • 67) 0.815 479 798 522 335 046 056 017 92 × 2 = 1 + 0.630 959 597 044 670 092 112 035 84;
  • 68) 0.630 959 597 044 670 092 112 035 84 × 2 = 1 + 0.261 919 194 089 340 184 224 071 68;
  • 69) 0.261 919 194 089 340 184 224 071 68 × 2 = 0 + 0.523 838 388 178 680 368 448 143 36;
  • 70) 0.523 838 388 178 680 368 448 143 36 × 2 = 1 + 0.047 676 776 357 360 736 896 286 72;
  • 71) 0.047 676 776 357 360 736 896 286 72 × 2 = 0 + 0.095 353 552 714 721 473 792 573 44;
  • 72) 0.095 353 552 714 721 473 792 573 44 × 2 = 0 + 0.190 707 105 429 442 947 585 146 88;
  • 73) 0.190 707 105 429 442 947 585 146 88 × 2 = 0 + 0.381 414 210 858 885 895 170 293 76;
  • 74) 0.381 414 210 858 885 895 170 293 76 × 2 = 0 + 0.762 828 421 717 771 790 340 587 52;
  • 75) 0.762 828 421 717 771 790 340 587 52 × 2 = 1 + 0.525 656 843 435 543 580 681 175 04;
  • 76) 0.525 656 843 435 543 580 681 175 04 × 2 = 1 + 0.051 313 686 871 087 161 362 350 08;
  • 77) 0.051 313 686 871 087 161 362 350 08 × 2 = 0 + 0.102 627 373 742 174 322 724 700 16;

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:


0.000 000 045 419 381 005 632 303 53(10) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2)

6. Positive number before normalization:

0.000 000 045 419 381 005 632 303 53(10) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 045 419 381 005 632 303 53(10) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2) =

0.0000 0000 0000 0000 0000 0000 1100 0011 0001 0011 0010 0011 0011 0101 1111 0101 0011 0100 0011 0(2) × 20 =

1.1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110(2) × 2-25


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): -25

Mantissa (not normalized):
1.1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(11-1) - 1 =

(-25 + 1 023)(10) =

998(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;
  • 998 ÷ 2 = 499 + 0;
  • 499 ÷ 2 = 249 + 1;
  • 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) =

998(10) =

011 1110 0110(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. 1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110 =

1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110


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 0110

Mantissa (52 bits) =
1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110


Decimal number -0.000 000 045 419 381 005 632 303 53 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 0110 - 1000 0110 0010 0110 0100 0110 0110 1011 1110 1010 0110 1000 0110

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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