1.595 242 277 440 550 187 010 811 934 502 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.595 242 277 440 550 187 010 811 934 502(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
1.595 242 277 440 550 187 010 811 934 502(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: 1.
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;
  • 1 ÷ 2 = 0 + 1;

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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.595 242 277 440 550 187 010 811 934 502.

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) 0.595 242 277 440 550 187 010 811 934 502 × 2 = 1 + 0.190 484 554 881 100 374 021 623 869 004;
  • 2) 0.190 484 554 881 100 374 021 623 869 004 × 2 = 0 + 0.380 969 109 762 200 748 043 247 738 008;
  • 3) 0.380 969 109 762 200 748 043 247 738 008 × 2 = 0 + 0.761 938 219 524 401 496 086 495 476 016;
  • 4) 0.761 938 219 524 401 496 086 495 476 016 × 2 = 1 + 0.523 876 439 048 802 992 172 990 952 032;
  • 5) 0.523 876 439 048 802 992 172 990 952 032 × 2 = 1 + 0.047 752 878 097 605 984 345 981 904 064;
  • 6) 0.047 752 878 097 605 984 345 981 904 064 × 2 = 0 + 0.095 505 756 195 211 968 691 963 808 128;
  • 7) 0.095 505 756 195 211 968 691 963 808 128 × 2 = 0 + 0.191 011 512 390 423 937 383 927 616 256;
  • 8) 0.191 011 512 390 423 937 383 927 616 256 × 2 = 0 + 0.382 023 024 780 847 874 767 855 232 512;
  • 9) 0.382 023 024 780 847 874 767 855 232 512 × 2 = 0 + 0.764 046 049 561 695 749 535 710 465 024;
  • 10) 0.764 046 049 561 695 749 535 710 465 024 × 2 = 1 + 0.528 092 099 123 391 499 071 420 930 048;
  • 11) 0.528 092 099 123 391 499 071 420 930 048 × 2 = 1 + 0.056 184 198 246 782 998 142 841 860 096;
  • 12) 0.056 184 198 246 782 998 142 841 860 096 × 2 = 0 + 0.112 368 396 493 565 996 285 683 720 192;
  • 13) 0.112 368 396 493 565 996 285 683 720 192 × 2 = 0 + 0.224 736 792 987 131 992 571 367 440 384;
  • 14) 0.224 736 792 987 131 992 571 367 440 384 × 2 = 0 + 0.449 473 585 974 263 985 142 734 880 768;
  • 15) 0.449 473 585 974 263 985 142 734 880 768 × 2 = 0 + 0.898 947 171 948 527 970 285 469 761 536;
  • 16) 0.898 947 171 948 527 970 285 469 761 536 × 2 = 1 + 0.797 894 343 897 055 940 570 939 523 072;
  • 17) 0.797 894 343 897 055 940 570 939 523 072 × 2 = 1 + 0.595 788 687 794 111 881 141 879 046 144;
  • 18) 0.595 788 687 794 111 881 141 879 046 144 × 2 = 1 + 0.191 577 375 588 223 762 283 758 092 288;
  • 19) 0.191 577 375 588 223 762 283 758 092 288 × 2 = 0 + 0.383 154 751 176 447 524 567 516 184 576;
  • 20) 0.383 154 751 176 447 524 567 516 184 576 × 2 = 0 + 0.766 309 502 352 895 049 135 032 369 152;
  • 21) 0.766 309 502 352 895 049 135 032 369 152 × 2 = 1 + 0.532 619 004 705 790 098 270 064 738 304;
  • 22) 0.532 619 004 705 790 098 270 064 738 304 × 2 = 1 + 0.065 238 009 411 580 196 540 129 476 608;
  • 23) 0.065 238 009 411 580 196 540 129 476 608 × 2 = 0 + 0.130 476 018 823 160 393 080 258 953 216;
  • 24) 0.130 476 018 823 160 393 080 258 953 216 × 2 = 0 + 0.260 952 037 646 320 786 160 517 906 432;
  • 25) 0.260 952 037 646 320 786 160 517 906 432 × 2 = 0 + 0.521 904 075 292 641 572 321 035 812 864;
  • 26) 0.521 904 075 292 641 572 321 035 812 864 × 2 = 1 + 0.043 808 150 585 283 144 642 071 625 728;
  • 27) 0.043 808 150 585 283 144 642 071 625 728 × 2 = 0 + 0.087 616 301 170 566 289 284 143 251 456;
  • 28) 0.087 616 301 170 566 289 284 143 251 456 × 2 = 0 + 0.175 232 602 341 132 578 568 286 502 912;
  • 29) 0.175 232 602 341 132 578 568 286 502 912 × 2 = 0 + 0.350 465 204 682 265 157 136 573 005 824;
  • 30) 0.350 465 204 682 265 157 136 573 005 824 × 2 = 0 + 0.700 930 409 364 530 314 273 146 011 648;
  • 31) 0.700 930 409 364 530 314 273 146 011 648 × 2 = 1 + 0.401 860 818 729 060 628 546 292 023 296;
  • 32) 0.401 860 818 729 060 628 546 292 023 296 × 2 = 0 + 0.803 721 637 458 121 257 092 584 046 592;
  • 33) 0.803 721 637 458 121 257 092 584 046 592 × 2 = 1 + 0.607 443 274 916 242 514 185 168 093 184;
  • 34) 0.607 443 274 916 242 514 185 168 093 184 × 2 = 1 + 0.214 886 549 832 485 028 370 336 186 368;
  • 35) 0.214 886 549 832 485 028 370 336 186 368 × 2 = 0 + 0.429 773 099 664 970 056 740 672 372 736;
  • 36) 0.429 773 099 664 970 056 740 672 372 736 × 2 = 0 + 0.859 546 199 329 940 113 481 344 745 472;
  • 37) 0.859 546 199 329 940 113 481 344 745 472 × 2 = 1 + 0.719 092 398 659 880 226 962 689 490 944;
  • 38) 0.719 092 398 659 880 226 962 689 490 944 × 2 = 1 + 0.438 184 797 319 760 453 925 378 981 888;
  • 39) 0.438 184 797 319 760 453 925 378 981 888 × 2 = 0 + 0.876 369 594 639 520 907 850 757 963 776;
  • 40) 0.876 369 594 639 520 907 850 757 963 776 × 2 = 1 + 0.752 739 189 279 041 815 701 515 927 552;
  • 41) 0.752 739 189 279 041 815 701 515 927 552 × 2 = 1 + 0.505 478 378 558 083 631 403 031 855 104;
  • 42) 0.505 478 378 558 083 631 403 031 855 104 × 2 = 1 + 0.010 956 757 116 167 262 806 063 710 208;
  • 43) 0.010 956 757 116 167 262 806 063 710 208 × 2 = 0 + 0.021 913 514 232 334 525 612 127 420 416;
  • 44) 0.021 913 514 232 334 525 612 127 420 416 × 2 = 0 + 0.043 827 028 464 669 051 224 254 840 832;
  • 45) 0.043 827 028 464 669 051 224 254 840 832 × 2 = 0 + 0.087 654 056 929 338 102 448 509 681 664;
  • 46) 0.087 654 056 929 338 102 448 509 681 664 × 2 = 0 + 0.175 308 113 858 676 204 897 019 363 328;
  • 47) 0.175 308 113 858 676 204 897 019 363 328 × 2 = 0 + 0.350 616 227 717 352 409 794 038 726 656;
  • 48) 0.350 616 227 717 352 409 794 038 726 656 × 2 = 0 + 0.701 232 455 434 704 819 588 077 453 312;
  • 49) 0.701 232 455 434 704 819 588 077 453 312 × 2 = 1 + 0.402 464 910 869 409 639 176 154 906 624;
  • 50) 0.402 464 910 869 409 639 176 154 906 624 × 2 = 0 + 0.804 929 821 738 819 278 352 309 813 248;
  • 51) 0.804 929 821 738 819 278 352 309 813 248 × 2 = 1 + 0.609 859 643 477 638 556 704 619 626 496;
  • 52) 0.609 859 643 477 638 556 704 619 626 496 × 2 = 1 + 0.219 719 286 955 277 113 409 239 252 992;
  • 53) 0.219 719 286 955 277 113 409 239 252 992 × 2 = 0 + 0.439 438 573 910 554 226 818 478 505 984;

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:


0.595 242 277 440 550 187 010 811 934 502(10) =

0.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2)

5. Positive number before normalization:

1.595 242 277 440 550 187 010 811 934 502(10) =

1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.595 242 277 440 550 187 010 811 934 502(10) =

1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2) =

1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2) × 20


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): 0

Mantissa (not normalized):
1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

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) =

1023(10) =

011 1111 1111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0 =

1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011


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 1111 1111

Mantissa (52 bits) =
1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011


Decimal number 1.595 242 277 440 550 187 010 811 934 502 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011

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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