32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 000 000 000 056 835 794 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 000 000 000 056 835 794(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 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: 0.000 000 000 000 056 835 794.

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) 0.000 000 000 000 056 835 794 × 2 = 0 + 0.000 000 000 000 113 671 588;
  • 2) 0.000 000 000 000 113 671 588 × 2 = 0 + 0.000 000 000 000 227 343 176;
  • 3) 0.000 000 000 000 227 343 176 × 2 = 0 + 0.000 000 000 000 454 686 352;
  • 4) 0.000 000 000 000 454 686 352 × 2 = 0 + 0.000 000 000 000 909 372 704;
  • 5) 0.000 000 000 000 909 372 704 × 2 = 0 + 0.000 000 000 001 818 745 408;
  • 6) 0.000 000 000 001 818 745 408 × 2 = 0 + 0.000 000 000 003 637 490 816;
  • 7) 0.000 000 000 003 637 490 816 × 2 = 0 + 0.000 000 000 007 274 981 632;
  • 8) 0.000 000 000 007 274 981 632 × 2 = 0 + 0.000 000 000 014 549 963 264;
  • 9) 0.000 000 000 014 549 963 264 × 2 = 0 + 0.000 000 000 029 099 926 528;
  • 10) 0.000 000 000 029 099 926 528 × 2 = 0 + 0.000 000 000 058 199 853 056;
  • 11) 0.000 000 000 058 199 853 056 × 2 = 0 + 0.000 000 000 116 399 706 112;
  • 12) 0.000 000 000 116 399 706 112 × 2 = 0 + 0.000 000 000 232 799 412 224;
  • 13) 0.000 000 000 232 799 412 224 × 2 = 0 + 0.000 000 000 465 598 824 448;
  • 14) 0.000 000 000 465 598 824 448 × 2 = 0 + 0.000 000 000 931 197 648 896;
  • 15) 0.000 000 000 931 197 648 896 × 2 = 0 + 0.000 000 001 862 395 297 792;
  • 16) 0.000 000 001 862 395 297 792 × 2 = 0 + 0.000 000 003 724 790 595 584;
  • 17) 0.000 000 003 724 790 595 584 × 2 = 0 + 0.000 000 007 449 581 191 168;
  • 18) 0.000 000 007 449 581 191 168 × 2 = 0 + 0.000 000 014 899 162 382 336;
  • 19) 0.000 000 014 899 162 382 336 × 2 = 0 + 0.000 000 029 798 324 764 672;
  • 20) 0.000 000 029 798 324 764 672 × 2 = 0 + 0.000 000 059 596 649 529 344;
  • 21) 0.000 000 059 596 649 529 344 × 2 = 0 + 0.000 000 119 193 299 058 688;
  • 22) 0.000 000 119 193 299 058 688 × 2 = 0 + 0.000 000 238 386 598 117 376;
  • 23) 0.000 000 238 386 598 117 376 × 2 = 0 + 0.000 000 476 773 196 234 752;
  • 24) 0.000 000 476 773 196 234 752 × 2 = 0 + 0.000 000 953 546 392 469 504;
  • 25) 0.000 000 953 546 392 469 504 × 2 = 0 + 0.000 001 907 092 784 939 008;
  • 26) 0.000 001 907 092 784 939 008 × 2 = 0 + 0.000 003 814 185 569 878 016;
  • 27) 0.000 003 814 185 569 878 016 × 2 = 0 + 0.000 007 628 371 139 756 032;
  • 28) 0.000 007 628 371 139 756 032 × 2 = 0 + 0.000 015 256 742 279 512 064;
  • 29) 0.000 015 256 742 279 512 064 × 2 = 0 + 0.000 030 513 484 559 024 128;
  • 30) 0.000 030 513 484 559 024 128 × 2 = 0 + 0.000 061 026 969 118 048 256;
  • 31) 0.000 061 026 969 118 048 256 × 2 = 0 + 0.000 122 053 938 236 096 512;
  • 32) 0.000 122 053 938 236 096 512 × 2 = 0 + 0.000 244 107 876 472 193 024;
  • 33) 0.000 244 107 876 472 193 024 × 2 = 0 + 0.000 488 215 752 944 386 048;
  • 34) 0.000 488 215 752 944 386 048 × 2 = 0 + 0.000 976 431 505 888 772 096;
  • 35) 0.000 976 431 505 888 772 096 × 2 = 0 + 0.001 952 863 011 777 544 192;
  • 36) 0.001 952 863 011 777 544 192 × 2 = 0 + 0.003 905 726 023 555 088 384;
  • 37) 0.003 905 726 023 555 088 384 × 2 = 0 + 0.007 811 452 047 110 176 768;
  • 38) 0.007 811 452 047 110 176 768 × 2 = 0 + 0.015 622 904 094 220 353 536;
  • 39) 0.015 622 904 094 220 353 536 × 2 = 0 + 0.031 245 808 188 440 707 072;
  • 40) 0.031 245 808 188 440 707 072 × 2 = 0 + 0.062 491 616 376 881 414 144;
  • 41) 0.062 491 616 376 881 414 144 × 2 = 0 + 0.124 983 232 753 762 828 288;
  • 42) 0.124 983 232 753 762 828 288 × 2 = 0 + 0.249 966 465 507 525 656 576;
  • 43) 0.249 966 465 507 525 656 576 × 2 = 0 + 0.499 932 931 015 051 313 152;
  • 44) 0.499 932 931 015 051 313 152 × 2 = 0 + 0.999 865 862 030 102 626 304;
  • 45) 0.999 865 862 030 102 626 304 × 2 = 1 + 0.999 731 724 060 205 252 608;
  • 46) 0.999 731 724 060 205 252 608 × 2 = 1 + 0.999 463 448 120 410 505 216;
  • 47) 0.999 463 448 120 410 505 216 × 2 = 1 + 0.998 926 896 240 821 010 432;
  • 48) 0.998 926 896 240 821 010 432 × 2 = 1 + 0.997 853 792 481 642 020 864;
  • 49) 0.997 853 792 481 642 020 864 × 2 = 1 + 0.995 707 584 963 284 041 728;
  • 50) 0.995 707 584 963 284 041 728 × 2 = 1 + 0.991 415 169 926 568 083 456;
  • 51) 0.991 415 169 926 568 083 456 × 2 = 1 + 0.982 830 339 853 136 166 912;
  • 52) 0.982 830 339 853 136 166 912 × 2 = 1 + 0.965 660 679 706 272 333 824;
  • 53) 0.965 660 679 706 272 333 824 × 2 = 1 + 0.931 321 359 412 544 667 648;
  • 54) 0.931 321 359 412 544 667 648 × 2 = 1 + 0.862 642 718 825 089 335 296;
  • 55) 0.862 642 718 825 089 335 296 × 2 = 1 + 0.725 285 437 650 178 670 592;
  • 56) 0.725 285 437 650 178 670 592 × 2 = 1 + 0.450 570 875 300 357 341 184;
  • 57) 0.450 570 875 300 357 341 184 × 2 = 0 + 0.901 141 750 600 714 682 368;
  • 58) 0.901 141 750 600 714 682 368 × 2 = 1 + 0.802 283 501 201 429 364 736;
  • 59) 0.802 283 501 201 429 364 736 × 2 = 1 + 0.604 567 002 402 858 729 472;
  • 60) 0.604 567 002 402 858 729 472 × 2 = 1 + 0.209 134 004 805 717 458 944;
  • 61) 0.209 134 004 805 717 458 944 × 2 = 0 + 0.418 268 009 611 434 917 888;
  • 62) 0.418 268 009 611 434 917 888 × 2 = 0 + 0.836 536 019 222 869 835 776;
  • 63) 0.836 536 019 222 869 835 776 × 2 = 1 + 0.673 072 038 445 739 671 552;
  • 64) 0.673 072 038 445 739 671 552 × 2 = 1 + 0.346 144 076 891 479 343 104;
  • 65) 0.346 144 076 891 479 343 104 × 2 = 0 + 0.692 288 153 782 958 686 208;
  • 66) 0.692 288 153 782 958 686 208 × 2 = 1 + 0.384 576 307 565 917 372 416;
  • 67) 0.384 576 307 565 917 372 416 × 2 = 0 + 0.769 152 615 131 834 744 832;
  • 68) 0.769 152 615 131 834 744 832 × 2 = 1 + 0.538 305 230 263 669 489 664;

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...)


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:


0.000 000 000 000 056 835 794(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 0111 0011 0101(2)


5. Positive number before normalization:

0.000 000 000 000 056 835 794(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 0111 0011 0101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 45 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 056 835 794(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 0111 0011 0101(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1111 1111 1111 0111 0011 0101(2) × 20 =

1.1111 1111 1110 1110 0110 101(2) × 2-45


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -45

Mantissa (not normalized):
1.1111 1111 1110 1110 0110 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-45 + 2(8-1) - 1 =

(-45 + 127)(10) =

82(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 82 ÷ 2 = 41 + 0;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

82(10) =

0101 0010(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 111 1111 1111 0111 0011 0101 =

111 1111 1111 0111 0011 0101


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0101 0010

Mantissa (23 bits) =
111 1111 1111 0111 0011 0101


The base ten decimal number 0.000 000 000 000 056 835 794 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0101 0010 - 111 1111 1111 0111 0011 0101

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number 0.000 000 000 000 056 835 793 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number 0.000 000 000 000 056 835 795 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111