32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 000 000 000 000 000 006 8 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 000 000 000 000 000 006 8(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: 0.000 000 000 000 000 000 006 8.

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) 0.000 000 000 000 000 000 006 8 × 2 = 0 + 0.000 000 000 000 000 000 013 6;
  • 2) 0.000 000 000 000 000 000 013 6 × 2 = 0 + 0.000 000 000 000 000 000 027 2;
  • 3) 0.000 000 000 000 000 000 027 2 × 2 = 0 + 0.000 000 000 000 000 000 054 4;
  • 4) 0.000 000 000 000 000 000 054 4 × 2 = 0 + 0.000 000 000 000 000 000 108 8;
  • 5) 0.000 000 000 000 000 000 108 8 × 2 = 0 + 0.000 000 000 000 000 000 217 6;
  • 6) 0.000 000 000 000 000 000 217 6 × 2 = 0 + 0.000 000 000 000 000 000 435 2;
  • 7) 0.000 000 000 000 000 000 435 2 × 2 = 0 + 0.000 000 000 000 000 000 870 4;
  • 8) 0.000 000 000 000 000 000 870 4 × 2 = 0 + 0.000 000 000 000 000 001 740 8;
  • 9) 0.000 000 000 000 000 001 740 8 × 2 = 0 + 0.000 000 000 000 000 003 481 6;
  • 10) 0.000 000 000 000 000 003 481 6 × 2 = 0 + 0.000 000 000 000 000 006 963 2;
  • 11) 0.000 000 000 000 000 006 963 2 × 2 = 0 + 0.000 000 000 000 000 013 926 4;
  • 12) 0.000 000 000 000 000 013 926 4 × 2 = 0 + 0.000 000 000 000 000 027 852 8;
  • 13) 0.000 000 000 000 000 027 852 8 × 2 = 0 + 0.000 000 000 000 000 055 705 6;
  • 14) 0.000 000 000 000 000 055 705 6 × 2 = 0 + 0.000 000 000 000 000 111 411 2;
  • 15) 0.000 000 000 000 000 111 411 2 × 2 = 0 + 0.000 000 000 000 000 222 822 4;
  • 16) 0.000 000 000 000 000 222 822 4 × 2 = 0 + 0.000 000 000 000 000 445 644 8;
  • 17) 0.000 000 000 000 000 445 644 8 × 2 = 0 + 0.000 000 000 000 000 891 289 6;
  • 18) 0.000 000 000 000 000 891 289 6 × 2 = 0 + 0.000 000 000 000 001 782 579 2;
  • 19) 0.000 000 000 000 001 782 579 2 × 2 = 0 + 0.000 000 000 000 003 565 158 4;
  • 20) 0.000 000 000 000 003 565 158 4 × 2 = 0 + 0.000 000 000 000 007 130 316 8;
  • 21) 0.000 000 000 000 007 130 316 8 × 2 = 0 + 0.000 000 000 000 014 260 633 6;
  • 22) 0.000 000 000 000 014 260 633 6 × 2 = 0 + 0.000 000 000 000 028 521 267 2;
  • 23) 0.000 000 000 000 028 521 267 2 × 2 = 0 + 0.000 000 000 000 057 042 534 4;
  • 24) 0.000 000 000 000 057 042 534 4 × 2 = 0 + 0.000 000 000 000 114 085 068 8;
  • 25) 0.000 000 000 000 114 085 068 8 × 2 = 0 + 0.000 000 000 000 228 170 137 6;
  • 26) 0.000 000 000 000 228 170 137 6 × 2 = 0 + 0.000 000 000 000 456 340 275 2;
  • 27) 0.000 000 000 000 456 340 275 2 × 2 = 0 + 0.000 000 000 000 912 680 550 4;
  • 28) 0.000 000 000 000 912 680 550 4 × 2 = 0 + 0.000 000 000 001 825 361 100 8;
  • 29) 0.000 000 000 001 825 361 100 8 × 2 = 0 + 0.000 000 000 003 650 722 201 6;
  • 30) 0.000 000 000 003 650 722 201 6 × 2 = 0 + 0.000 000 000 007 301 444 403 2;
  • 31) 0.000 000 000 007 301 444 403 2 × 2 = 0 + 0.000 000 000 014 602 888 806 4;
  • 32) 0.000 000 000 014 602 888 806 4 × 2 = 0 + 0.000 000 000 029 205 777 612 8;
  • 33) 0.000 000 000 029 205 777 612 8 × 2 = 0 + 0.000 000 000 058 411 555 225 6;
  • 34) 0.000 000 000 058 411 555 225 6 × 2 = 0 + 0.000 000 000 116 823 110 451 2;
  • 35) 0.000 000 000 116 823 110 451 2 × 2 = 0 + 0.000 000 000 233 646 220 902 4;
  • 36) 0.000 000 000 233 646 220 902 4 × 2 = 0 + 0.000 000 000 467 292 441 804 8;
  • 37) 0.000 000 000 467 292 441 804 8 × 2 = 0 + 0.000 000 000 934 584 883 609 6;
  • 38) 0.000 000 000 934 584 883 609 6 × 2 = 0 + 0.000 000 001 869 169 767 219 2;
  • 39) 0.000 000 001 869 169 767 219 2 × 2 = 0 + 0.000 000 003 738 339 534 438 4;
  • 40) 0.000 000 003 738 339 534 438 4 × 2 = 0 + 0.000 000 007 476 679 068 876 8;
  • 41) 0.000 000 007 476 679 068 876 8 × 2 = 0 + 0.000 000 014 953 358 137 753 6;
  • 42) 0.000 000 014 953 358 137 753 6 × 2 = 0 + 0.000 000 029 906 716 275 507 2;
  • 43) 0.000 000 029 906 716 275 507 2 × 2 = 0 + 0.000 000 059 813 432 551 014 4;
  • 44) 0.000 000 059 813 432 551 014 4 × 2 = 0 + 0.000 000 119 626 865 102 028 8;
  • 45) 0.000 000 119 626 865 102 028 8 × 2 = 0 + 0.000 000 239 253 730 204 057 6;
  • 46) 0.000 000 239 253 730 204 057 6 × 2 = 0 + 0.000 000 478 507 460 408 115 2;
  • 47) 0.000 000 478 507 460 408 115 2 × 2 = 0 + 0.000 000 957 014 920 816 230 4;
  • 48) 0.000 000 957 014 920 816 230 4 × 2 = 0 + 0.000 001 914 029 841 632 460 8;
  • 49) 0.000 001 914 029 841 632 460 8 × 2 = 0 + 0.000 003 828 059 683 264 921 6;
  • 50) 0.000 003 828 059 683 264 921 6 × 2 = 0 + 0.000 007 656 119 366 529 843 2;
  • 51) 0.000 007 656 119 366 529 843 2 × 2 = 0 + 0.000 015 312 238 733 059 686 4;
  • 52) 0.000 015 312 238 733 059 686 4 × 2 = 0 + 0.000 030 624 477 466 119 372 8;
  • 53) 0.000 030 624 477 466 119 372 8 × 2 = 0 + 0.000 061 248 954 932 238 745 6;
  • 54) 0.000 061 248 954 932 238 745 6 × 2 = 0 + 0.000 122 497 909 864 477 491 2;
  • 55) 0.000 122 497 909 864 477 491 2 × 2 = 0 + 0.000 244 995 819 728 954 982 4;
  • 56) 0.000 244 995 819 728 954 982 4 × 2 = 0 + 0.000 489 991 639 457 909 964 8;
  • 57) 0.000 489 991 639 457 909 964 8 × 2 = 0 + 0.000 979 983 278 915 819 929 6;
  • 58) 0.000 979 983 278 915 819 929 6 × 2 = 0 + 0.001 959 966 557 831 639 859 2;
  • 59) 0.001 959 966 557 831 639 859 2 × 2 = 0 + 0.003 919 933 115 663 279 718 4;
  • 60) 0.003 919 933 115 663 279 718 4 × 2 = 0 + 0.007 839 866 231 326 559 436 8;
  • 61) 0.007 839 866 231 326 559 436 8 × 2 = 0 + 0.015 679 732 462 653 118 873 6;
  • 62) 0.015 679 732 462 653 118 873 6 × 2 = 0 + 0.031 359 464 925 306 237 747 2;
  • 63) 0.031 359 464 925 306 237 747 2 × 2 = 0 + 0.062 718 929 850 612 475 494 4;
  • 64) 0.062 718 929 850 612 475 494 4 × 2 = 0 + 0.125 437 859 701 224 950 988 8;
  • 65) 0.125 437 859 701 224 950 988 8 × 2 = 0 + 0.250 875 719 402 449 901 977 6;
  • 66) 0.250 875 719 402 449 901 977 6 × 2 = 0 + 0.501 751 438 804 899 803 955 2;
  • 67) 0.501 751 438 804 899 803 955 2 × 2 = 1 + 0.003 502 877 609 799 607 910 4;
  • 68) 0.003 502 877 609 799 607 910 4 × 2 = 0 + 0.007 005 755 219 599 215 820 8;
  • 69) 0.007 005 755 219 599 215 820 8 × 2 = 0 + 0.014 011 510 439 198 431 641 6;
  • 70) 0.014 011 510 439 198 431 641 6 × 2 = 0 + 0.028 023 020 878 396 863 283 2;
  • 71) 0.028 023 020 878 396 863 283 2 × 2 = 0 + 0.056 046 041 756 793 726 566 4;
  • 72) 0.056 046 041 756 793 726 566 4 × 2 = 0 + 0.112 092 083 513 587 453 132 8;
  • 73) 0.112 092 083 513 587 453 132 8 × 2 = 0 + 0.224 184 167 027 174 906 265 6;
  • 74) 0.224 184 167 027 174 906 265 6 × 2 = 0 + 0.448 368 334 054 349 812 531 2;
  • 75) 0.448 368 334 054 349 812 531 2 × 2 = 0 + 0.896 736 668 108 699 625 062 4;
  • 76) 0.896 736 668 108 699 625 062 4 × 2 = 1 + 0.793 473 336 217 399 250 124 8;
  • 77) 0.793 473 336 217 399 250 124 8 × 2 = 1 + 0.586 946 672 434 798 500 249 6;
  • 78) 0.586 946 672 434 798 500 249 6 × 2 = 1 + 0.173 893 344 869 597 000 499 2;
  • 79) 0.173 893 344 869 597 000 499 2 × 2 = 0 + 0.347 786 689 739 194 000 998 4;
  • 80) 0.347 786 689 739 194 000 998 4 × 2 = 0 + 0.695 573 379 478 388 001 996 8;
  • 81) 0.695 573 379 478 388 001 996 8 × 2 = 1 + 0.391 146 758 956 776 003 993 6;
  • 82) 0.391 146 758 956 776 003 993 6 × 2 = 0 + 0.782 293 517 913 552 007 987 2;
  • 83) 0.782 293 517 913 552 007 987 2 × 2 = 1 + 0.564 587 035 827 104 015 974 4;
  • 84) 0.564 587 035 827 104 015 974 4 × 2 = 1 + 0.129 174 071 654 208 031 948 8;
  • 85) 0.129 174 071 654 208 031 948 8 × 2 = 0 + 0.258 348 143 308 416 063 897 6;
  • 86) 0.258 348 143 308 416 063 897 6 × 2 = 0 + 0.516 696 286 616 832 127 795 2;
  • 87) 0.516 696 286 616 832 127 795 2 × 2 = 1 + 0.033 392 573 233 664 255 590 4;
  • 88) 0.033 392 573 233 664 255 590 4 × 2 = 0 + 0.066 785 146 467 328 511 180 8;
  • 89) 0.066 785 146 467 328 511 180 8 × 2 = 0 + 0.133 570 292 934 657 022 361 6;
  • 90) 0.133 570 292 934 657 022 361 6 × 2 = 0 + 0.267 140 585 869 314 044 723 2;

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:


0.000 000 000 000 000 000 006 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0001 1100 1011 0010 00(2)


5. Positive number before normalization:

0.000 000 000 000 000 000 006 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0001 1100 1011 0010 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 67 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 006 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0001 1100 1011 0010 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000 0001 1100 1011 0010 00(2) × 20 =

1.0000 0000 1110 0101 1001 000(2) × 2-67


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): -67

Mantissa (not normalized):
1.0000 0000 1110 0101 1001 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-67 + 2(8-1) - 1 =

(-67 + 127)(10) =

60(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;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 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) =

60(10) =

0011 1100(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. 000 0000 0111 0010 1100 1000 =

000 0000 0111 0010 1100 1000


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) =
0011 1100

Mantissa (23 bits) =
000 0000 0111 0010 1100 1000


The base ten decimal number 0.000 000 000 000 000 000 006 8 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0011 1100 - 000 0000 0111 0010 1100 1000

More operations with decimal numbers converted 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