1.745 459 324 169 999 826 281 696 186 557 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 557(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
1.745 459 324 169 999 826 281 696 186 557(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: 0.745 459 324 169 999 826 281 696 186 557.

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) 0.745 459 324 169 999 826 281 696 186 557 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 114;
  • 2) 0.490 918 648 339 999 652 563 392 373 114 × 2 = 0 + 0.981 837 296 679 999 305 126 784 746 228;
  • 3) 0.981 837 296 679 999 305 126 784 746 228 × 2 = 1 + 0.963 674 593 359 998 610 253 569 492 456;
  • 4) 0.963 674 593 359 998 610 253 569 492 456 × 2 = 1 + 0.927 349 186 719 997 220 507 138 984 912;
  • 5) 0.927 349 186 719 997 220 507 138 984 912 × 2 = 1 + 0.854 698 373 439 994 441 014 277 969 824;
  • 6) 0.854 698 373 439 994 441 014 277 969 824 × 2 = 1 + 0.709 396 746 879 988 882 028 555 939 648;
  • 7) 0.709 396 746 879 988 882 028 555 939 648 × 2 = 1 + 0.418 793 493 759 977 764 057 111 879 296;
  • 8) 0.418 793 493 759 977 764 057 111 879 296 × 2 = 0 + 0.837 586 987 519 955 528 114 223 758 592;
  • 9) 0.837 586 987 519 955 528 114 223 758 592 × 2 = 1 + 0.675 173 975 039 911 056 228 447 517 184;
  • 10) 0.675 173 975 039 911 056 228 447 517 184 × 2 = 1 + 0.350 347 950 079 822 112 456 895 034 368;
  • 11) 0.350 347 950 079 822 112 456 895 034 368 × 2 = 0 + 0.700 695 900 159 644 224 913 790 068 736;
  • 12) 0.700 695 900 159 644 224 913 790 068 736 × 2 = 1 + 0.401 391 800 319 288 449 827 580 137 472;
  • 13) 0.401 391 800 319 288 449 827 580 137 472 × 2 = 0 + 0.802 783 600 638 576 899 655 160 274 944;
  • 14) 0.802 783 600 638 576 899 655 160 274 944 × 2 = 1 + 0.605 567 201 277 153 799 310 320 549 888;
  • 15) 0.605 567 201 277 153 799 310 320 549 888 × 2 = 1 + 0.211 134 402 554 307 598 620 641 099 776;
  • 16) 0.211 134 402 554 307 598 620 641 099 776 × 2 = 0 + 0.422 268 805 108 615 197 241 282 199 552;
  • 17) 0.422 268 805 108 615 197 241 282 199 552 × 2 = 0 + 0.844 537 610 217 230 394 482 564 399 104;
  • 18) 0.844 537 610 217 230 394 482 564 399 104 × 2 = 1 + 0.689 075 220 434 460 788 965 128 798 208;
  • 19) 0.689 075 220 434 460 788 965 128 798 208 × 2 = 1 + 0.378 150 440 868 921 577 930 257 596 416;
  • 20) 0.378 150 440 868 921 577 930 257 596 416 × 2 = 0 + 0.756 300 881 737 843 155 860 515 192 832;
  • 21) 0.756 300 881 737 843 155 860 515 192 832 × 2 = 1 + 0.512 601 763 475 686 311 721 030 385 664;
  • 22) 0.512 601 763 475 686 311 721 030 385 664 × 2 = 1 + 0.025 203 526 951 372 623 442 060 771 328;
  • 23) 0.025 203 526 951 372 623 442 060 771 328 × 2 = 0 + 0.050 407 053 902 745 246 884 121 542 656;
  • 24) 0.050 407 053 902 745 246 884 121 542 656 × 2 = 0 + 0.100 814 107 805 490 493 768 243 085 312;
  • 25) 0.100 814 107 805 490 493 768 243 085 312 × 2 = 0 + 0.201 628 215 610 980 987 536 486 170 624;
  • 26) 0.201 628 215 610 980 987 536 486 170 624 × 2 = 0 + 0.403 256 431 221 961 975 072 972 341 248;
  • 27) 0.403 256 431 221 961 975 072 972 341 248 × 2 = 0 + 0.806 512 862 443 923 950 145 944 682 496;
  • 28) 0.806 512 862 443 923 950 145 944 682 496 × 2 = 1 + 0.613 025 724 887 847 900 291 889 364 992;
  • 29) 0.613 025 724 887 847 900 291 889 364 992 × 2 = 1 + 0.226 051 449 775 695 800 583 778 729 984;
  • 30) 0.226 051 449 775 695 800 583 778 729 984 × 2 = 0 + 0.452 102 899 551 391 601 167 557 459 968;
  • 31) 0.452 102 899 551 391 601 167 557 459 968 × 2 = 0 + 0.904 205 799 102 783 202 335 114 919 936;
  • 32) 0.904 205 799 102 783 202 335 114 919 936 × 2 = 1 + 0.808 411 598 205 566 404 670 229 839 872;
  • 33) 0.808 411 598 205 566 404 670 229 839 872 × 2 = 1 + 0.616 823 196 411 132 809 340 459 679 744;
  • 34) 0.616 823 196 411 132 809 340 459 679 744 × 2 = 1 + 0.233 646 392 822 265 618 680 919 359 488;
  • 35) 0.233 646 392 822 265 618 680 919 359 488 × 2 = 0 + 0.467 292 785 644 531 237 361 838 718 976;
  • 36) 0.467 292 785 644 531 237 361 838 718 976 × 2 = 0 + 0.934 585 571 289 062 474 723 677 437 952;
  • 37) 0.934 585 571 289 062 474 723 677 437 952 × 2 = 1 + 0.869 171 142 578 124 949 447 354 875 904;
  • 38) 0.869 171 142 578 124 949 447 354 875 904 × 2 = 1 + 0.738 342 285 156 249 898 894 709 751 808;
  • 39) 0.738 342 285 156 249 898 894 709 751 808 × 2 = 1 + 0.476 684 570 312 499 797 789 419 503 616;
  • 40) 0.476 684 570 312 499 797 789 419 503 616 × 2 = 0 + 0.953 369 140 624 999 595 578 839 007 232;
  • 41) 0.953 369 140 624 999 595 578 839 007 232 × 2 = 1 + 0.906 738 281 249 999 191 157 678 014 464;
  • 42) 0.906 738 281 249 999 191 157 678 014 464 × 2 = 1 + 0.813 476 562 499 998 382 315 356 028 928;
  • 43) 0.813 476 562 499 998 382 315 356 028 928 × 2 = 1 + 0.626 953 124 999 996 764 630 712 057 856;
  • 44) 0.626 953 124 999 996 764 630 712 057 856 × 2 = 1 + 0.253 906 249 999 993 529 261 424 115 712;
  • 45) 0.253 906 249 999 993 529 261 424 115 712 × 2 = 0 + 0.507 812 499 999 987 058 522 848 231 424;
  • 46) 0.507 812 499 999 987 058 522 848 231 424 × 2 = 1 + 0.015 624 999 999 974 117 045 696 462 848;
  • 47) 0.015 624 999 999 974 117 045 696 462 848 × 2 = 0 + 0.031 249 999 999 948 234 091 392 925 696;
  • 48) 0.031 249 999 999 948 234 091 392 925 696 × 2 = 0 + 0.062 499 999 999 896 468 182 785 851 392;
  • 49) 0.062 499 999 999 896 468 182 785 851 392 × 2 = 0 + 0.124 999 999 999 792 936 365 571 702 784;
  • 50) 0.124 999 999 999 792 936 365 571 702 784 × 2 = 0 + 0.249 999 999 999 585 872 731 143 405 568;
  • 51) 0.249 999 999 999 585 872 731 143 405 568 × 2 = 0 + 0.499 999 999 999 171 745 462 286 811 136;
  • 52) 0.499 999 999 999 171 745 462 286 811 136 × 2 = 0 + 0.999 999 999 998 343 490 924 573 622 272;
  • 53) 0.999 999 999 998 343 490 924 573 622 272 × 2 = 1 + 0.999 999 999 996 686 981 849 147 244 544;

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:


0.745 459 324 169 999 826 281 696 186 557(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 557(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:


1.745 459 324 169 999 826 281 696 186 557(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 1.745 459 324 169 999 826 281 696 186 557 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