1.745 459 324 169 999 826 281 695 977 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 695 977(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
1.745 459 324 169 999 826 281 695 977(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: 0.745 459 324 169 999 826 281 695 977.

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) 0.745 459 324 169 999 826 281 695 977 × 2 = 1 + 0.490 918 648 339 999 652 563 391 954;
  • 2) 0.490 918 648 339 999 652 563 391 954 × 2 = 0 + 0.981 837 296 679 999 305 126 783 908;
  • 3) 0.981 837 296 679 999 305 126 783 908 × 2 = 1 + 0.963 674 593 359 998 610 253 567 816;
  • 4) 0.963 674 593 359 998 610 253 567 816 × 2 = 1 + 0.927 349 186 719 997 220 507 135 632;
  • 5) 0.927 349 186 719 997 220 507 135 632 × 2 = 1 + 0.854 698 373 439 994 441 014 271 264;
  • 6) 0.854 698 373 439 994 441 014 271 264 × 2 = 1 + 0.709 396 746 879 988 882 028 542 528;
  • 7) 0.709 396 746 879 988 882 028 542 528 × 2 = 1 + 0.418 793 493 759 977 764 057 085 056;
  • 8) 0.418 793 493 759 977 764 057 085 056 × 2 = 0 + 0.837 586 987 519 955 528 114 170 112;
  • 9) 0.837 586 987 519 955 528 114 170 112 × 2 = 1 + 0.675 173 975 039 911 056 228 340 224;
  • 10) 0.675 173 975 039 911 056 228 340 224 × 2 = 1 + 0.350 347 950 079 822 112 456 680 448;
  • 11) 0.350 347 950 079 822 112 456 680 448 × 2 = 0 + 0.700 695 900 159 644 224 913 360 896;
  • 12) 0.700 695 900 159 644 224 913 360 896 × 2 = 1 + 0.401 391 800 319 288 449 826 721 792;
  • 13) 0.401 391 800 319 288 449 826 721 792 × 2 = 0 + 0.802 783 600 638 576 899 653 443 584;
  • 14) 0.802 783 600 638 576 899 653 443 584 × 2 = 1 + 0.605 567 201 277 153 799 306 887 168;
  • 15) 0.605 567 201 277 153 799 306 887 168 × 2 = 1 + 0.211 134 402 554 307 598 613 774 336;
  • 16) 0.211 134 402 554 307 598 613 774 336 × 2 = 0 + 0.422 268 805 108 615 197 227 548 672;
  • 17) 0.422 268 805 108 615 197 227 548 672 × 2 = 0 + 0.844 537 610 217 230 394 455 097 344;
  • 18) 0.844 537 610 217 230 394 455 097 344 × 2 = 1 + 0.689 075 220 434 460 788 910 194 688;
  • 19) 0.689 075 220 434 460 788 910 194 688 × 2 = 1 + 0.378 150 440 868 921 577 820 389 376;
  • 20) 0.378 150 440 868 921 577 820 389 376 × 2 = 0 + 0.756 300 881 737 843 155 640 778 752;
  • 21) 0.756 300 881 737 843 155 640 778 752 × 2 = 1 + 0.512 601 763 475 686 311 281 557 504;
  • 22) 0.512 601 763 475 686 311 281 557 504 × 2 = 1 + 0.025 203 526 951 372 622 563 115 008;
  • 23) 0.025 203 526 951 372 622 563 115 008 × 2 = 0 + 0.050 407 053 902 745 245 126 230 016;
  • 24) 0.050 407 053 902 745 245 126 230 016 × 2 = 0 + 0.100 814 107 805 490 490 252 460 032;
  • 25) 0.100 814 107 805 490 490 252 460 032 × 2 = 0 + 0.201 628 215 610 980 980 504 920 064;
  • 26) 0.201 628 215 610 980 980 504 920 064 × 2 = 0 + 0.403 256 431 221 961 961 009 840 128;
  • 27) 0.403 256 431 221 961 961 009 840 128 × 2 = 0 + 0.806 512 862 443 923 922 019 680 256;
  • 28) 0.806 512 862 443 923 922 019 680 256 × 2 = 1 + 0.613 025 724 887 847 844 039 360 512;
  • 29) 0.613 025 724 887 847 844 039 360 512 × 2 = 1 + 0.226 051 449 775 695 688 078 721 024;
  • 30) 0.226 051 449 775 695 688 078 721 024 × 2 = 0 + 0.452 102 899 551 391 376 157 442 048;
  • 31) 0.452 102 899 551 391 376 157 442 048 × 2 = 0 + 0.904 205 799 102 782 752 314 884 096;
  • 32) 0.904 205 799 102 782 752 314 884 096 × 2 = 1 + 0.808 411 598 205 565 504 629 768 192;
  • 33) 0.808 411 598 205 565 504 629 768 192 × 2 = 1 + 0.616 823 196 411 131 009 259 536 384;
  • 34) 0.616 823 196 411 131 009 259 536 384 × 2 = 1 + 0.233 646 392 822 262 018 519 072 768;
  • 35) 0.233 646 392 822 262 018 519 072 768 × 2 = 0 + 0.467 292 785 644 524 037 038 145 536;
  • 36) 0.467 292 785 644 524 037 038 145 536 × 2 = 0 + 0.934 585 571 289 048 074 076 291 072;
  • 37) 0.934 585 571 289 048 074 076 291 072 × 2 = 1 + 0.869 171 142 578 096 148 152 582 144;
  • 38) 0.869 171 142 578 096 148 152 582 144 × 2 = 1 + 0.738 342 285 156 192 296 305 164 288;
  • 39) 0.738 342 285 156 192 296 305 164 288 × 2 = 1 + 0.476 684 570 312 384 592 610 328 576;
  • 40) 0.476 684 570 312 384 592 610 328 576 × 2 = 0 + 0.953 369 140 624 769 185 220 657 152;
  • 41) 0.953 369 140 624 769 185 220 657 152 × 2 = 1 + 0.906 738 281 249 538 370 441 314 304;
  • 42) 0.906 738 281 249 538 370 441 314 304 × 2 = 1 + 0.813 476 562 499 076 740 882 628 608;
  • 43) 0.813 476 562 499 076 740 882 628 608 × 2 = 1 + 0.626 953 124 998 153 481 765 257 216;
  • 44) 0.626 953 124 998 153 481 765 257 216 × 2 = 1 + 0.253 906 249 996 306 963 530 514 432;
  • 45) 0.253 906 249 996 306 963 530 514 432 × 2 = 0 + 0.507 812 499 992 613 927 061 028 864;
  • 46) 0.507 812 499 992 613 927 061 028 864 × 2 = 1 + 0.015 624 999 985 227 854 122 057 728;
  • 47) 0.015 624 999 985 227 854 122 057 728 × 2 = 0 + 0.031 249 999 970 455 708 244 115 456;
  • 48) 0.031 249 999 970 455 708 244 115 456 × 2 = 0 + 0.062 499 999 940 911 416 488 230 912;
  • 49) 0.062 499 999 940 911 416 488 230 912 × 2 = 0 + 0.124 999 999 881 822 832 976 461 824;
  • 50) 0.124 999 999 881 822 832 976 461 824 × 2 = 0 + 0.249 999 999 763 645 665 952 923 648;
  • 51) 0.249 999 999 763 645 665 952 923 648 × 2 = 0 + 0.499 999 999 527 291 331 905 847 296;
  • 52) 0.499 999 999 527 291 331 905 847 296 × 2 = 0 + 0.999 999 999 054 582 663 811 694 592;
  • 53) 0.999 999 999 054 582 663 811 694 592 × 2 = 1 + 0.999 999 998 109 165 327 623 389 184;

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:


0.745 459 324 169 999 826 281 695 977(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 695 977(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(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:


1.745 459 324 169 999 826 281 695 977(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(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):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1


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. 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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) =
1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


Decimal number 1.745 459 324 169 999 826 281 695 977 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000

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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