32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -0.000 000 000 000 000 002 12 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.000 000 000 000 000 002 12(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. Start with the positive version of the number:

|-0.000 000 000 000 000 002 12| = 0.000 000 000 000 000 002 12

2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 000 000 002 12.

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) 0.000 000 000 000 000 002 12 × 2 = 0 + 0.000 000 000 000 000 004 24;
  • 2) 0.000 000 000 000 000 004 24 × 2 = 0 + 0.000 000 000 000 000 008 48;
  • 3) 0.000 000 000 000 000 008 48 × 2 = 0 + 0.000 000 000 000 000 016 96;
  • 4) 0.000 000 000 000 000 016 96 × 2 = 0 + 0.000 000 000 000 000 033 92;
  • 5) 0.000 000 000 000 000 033 92 × 2 = 0 + 0.000 000 000 000 000 067 84;
  • 6) 0.000 000 000 000 000 067 84 × 2 = 0 + 0.000 000 000 000 000 135 68;
  • 7) 0.000 000 000 000 000 135 68 × 2 = 0 + 0.000 000 000 000 000 271 36;
  • 8) 0.000 000 000 000 000 271 36 × 2 = 0 + 0.000 000 000 000 000 542 72;
  • 9) 0.000 000 000 000 000 542 72 × 2 = 0 + 0.000 000 000 000 001 085 44;
  • 10) 0.000 000 000 000 001 085 44 × 2 = 0 + 0.000 000 000 000 002 170 88;
  • 11) 0.000 000 000 000 002 170 88 × 2 = 0 + 0.000 000 000 000 004 341 76;
  • 12) 0.000 000 000 000 004 341 76 × 2 = 0 + 0.000 000 000 000 008 683 52;
  • 13) 0.000 000 000 000 008 683 52 × 2 = 0 + 0.000 000 000 000 017 367 04;
  • 14) 0.000 000 000 000 017 367 04 × 2 = 0 + 0.000 000 000 000 034 734 08;
  • 15) 0.000 000 000 000 034 734 08 × 2 = 0 + 0.000 000 000 000 069 468 16;
  • 16) 0.000 000 000 000 069 468 16 × 2 = 0 + 0.000 000 000 000 138 936 32;
  • 17) 0.000 000 000 000 138 936 32 × 2 = 0 + 0.000 000 000 000 277 872 64;
  • 18) 0.000 000 000 000 277 872 64 × 2 = 0 + 0.000 000 000 000 555 745 28;
  • 19) 0.000 000 000 000 555 745 28 × 2 = 0 + 0.000 000 000 001 111 490 56;
  • 20) 0.000 000 000 001 111 490 56 × 2 = 0 + 0.000 000 000 002 222 981 12;
  • 21) 0.000 000 000 002 222 981 12 × 2 = 0 + 0.000 000 000 004 445 962 24;
  • 22) 0.000 000 000 004 445 962 24 × 2 = 0 + 0.000 000 000 008 891 924 48;
  • 23) 0.000 000 000 008 891 924 48 × 2 = 0 + 0.000 000 000 017 783 848 96;
  • 24) 0.000 000 000 017 783 848 96 × 2 = 0 + 0.000 000 000 035 567 697 92;
  • 25) 0.000 000 000 035 567 697 92 × 2 = 0 + 0.000 000 000 071 135 395 84;
  • 26) 0.000 000 000 071 135 395 84 × 2 = 0 + 0.000 000 000 142 270 791 68;
  • 27) 0.000 000 000 142 270 791 68 × 2 = 0 + 0.000 000 000 284 541 583 36;
  • 28) 0.000 000 000 284 541 583 36 × 2 = 0 + 0.000 000 000 569 083 166 72;
  • 29) 0.000 000 000 569 083 166 72 × 2 = 0 + 0.000 000 001 138 166 333 44;
  • 30) 0.000 000 001 138 166 333 44 × 2 = 0 + 0.000 000 002 276 332 666 88;
  • 31) 0.000 000 002 276 332 666 88 × 2 = 0 + 0.000 000 004 552 665 333 76;
  • 32) 0.000 000 004 552 665 333 76 × 2 = 0 + 0.000 000 009 105 330 667 52;
  • 33) 0.000 000 009 105 330 667 52 × 2 = 0 + 0.000 000 018 210 661 335 04;
  • 34) 0.000 000 018 210 661 335 04 × 2 = 0 + 0.000 000 036 421 322 670 08;
  • 35) 0.000 000 036 421 322 670 08 × 2 = 0 + 0.000 000 072 842 645 340 16;
  • 36) 0.000 000 072 842 645 340 16 × 2 = 0 + 0.000 000 145 685 290 680 32;
  • 37) 0.000 000 145 685 290 680 32 × 2 = 0 + 0.000 000 291 370 581 360 64;
  • 38) 0.000 000 291 370 581 360 64 × 2 = 0 + 0.000 000 582 741 162 721 28;
  • 39) 0.000 000 582 741 162 721 28 × 2 = 0 + 0.000 001 165 482 325 442 56;
  • 40) 0.000 001 165 482 325 442 56 × 2 = 0 + 0.000 002 330 964 650 885 12;
  • 41) 0.000 002 330 964 650 885 12 × 2 = 0 + 0.000 004 661 929 301 770 24;
  • 42) 0.000 004 661 929 301 770 24 × 2 = 0 + 0.000 009 323 858 603 540 48;
  • 43) 0.000 009 323 858 603 540 48 × 2 = 0 + 0.000 018 647 717 207 080 96;
  • 44) 0.000 018 647 717 207 080 96 × 2 = 0 + 0.000 037 295 434 414 161 92;
  • 45) 0.000 037 295 434 414 161 92 × 2 = 0 + 0.000 074 590 868 828 323 84;
  • 46) 0.000 074 590 868 828 323 84 × 2 = 0 + 0.000 149 181 737 656 647 68;
  • 47) 0.000 149 181 737 656 647 68 × 2 = 0 + 0.000 298 363 475 313 295 36;
  • 48) 0.000 298 363 475 313 295 36 × 2 = 0 + 0.000 596 726 950 626 590 72;
  • 49) 0.000 596 726 950 626 590 72 × 2 = 0 + 0.001 193 453 901 253 181 44;
  • 50) 0.001 193 453 901 253 181 44 × 2 = 0 + 0.002 386 907 802 506 362 88;
  • 51) 0.002 386 907 802 506 362 88 × 2 = 0 + 0.004 773 815 605 012 725 76;
  • 52) 0.004 773 815 605 012 725 76 × 2 = 0 + 0.009 547 631 210 025 451 52;
  • 53) 0.009 547 631 210 025 451 52 × 2 = 0 + 0.019 095 262 420 050 903 04;
  • 54) 0.019 095 262 420 050 903 04 × 2 = 0 + 0.038 190 524 840 101 806 08;
  • 55) 0.038 190 524 840 101 806 08 × 2 = 0 + 0.076 381 049 680 203 612 16;
  • 56) 0.076 381 049 680 203 612 16 × 2 = 0 + 0.152 762 099 360 407 224 32;
  • 57) 0.152 762 099 360 407 224 32 × 2 = 0 + 0.305 524 198 720 814 448 64;
  • 58) 0.305 524 198 720 814 448 64 × 2 = 0 + 0.611 048 397 441 628 897 28;
  • 59) 0.611 048 397 441 628 897 28 × 2 = 1 + 0.222 096 794 883 257 794 56;
  • 60) 0.222 096 794 883 257 794 56 × 2 = 0 + 0.444 193 589 766 515 589 12;
  • 61) 0.444 193 589 766 515 589 12 × 2 = 0 + 0.888 387 179 533 031 178 24;
  • 62) 0.888 387 179 533 031 178 24 × 2 = 1 + 0.776 774 359 066 062 356 48;
  • 63) 0.776 774 359 066 062 356 48 × 2 = 1 + 0.553 548 718 132 124 712 96;
  • 64) 0.553 548 718 132 124 712 96 × 2 = 1 + 0.107 097 436 264 249 425 92;
  • 65) 0.107 097 436 264 249 425 92 × 2 = 0 + 0.214 194 872 528 498 851 84;
  • 66) 0.214 194 872 528 498 851 84 × 2 = 0 + 0.428 389 745 056 997 703 68;
  • 67) 0.428 389 745 056 997 703 68 × 2 = 0 + 0.856 779 490 113 995 407 36;
  • 68) 0.856 779 490 113 995 407 36 × 2 = 1 + 0.713 558 980 227 990 814 72;
  • 69) 0.713 558 980 227 990 814 72 × 2 = 1 + 0.427 117 960 455 981 629 44;
  • 70) 0.427 117 960 455 981 629 44 × 2 = 0 + 0.854 235 920 911 963 258 88;
  • 71) 0.854 235 920 911 963 258 88 × 2 = 1 + 0.708 471 841 823 926 517 76;
  • 72) 0.708 471 841 823 926 517 76 × 2 = 1 + 0.416 943 683 647 853 035 52;
  • 73) 0.416 943 683 647 853 035 52 × 2 = 0 + 0.833 887 367 295 706 071 04;
  • 74) 0.833 887 367 295 706 071 04 × 2 = 1 + 0.667 774 734 591 412 142 08;
  • 75) 0.667 774 734 591 412 142 08 × 2 = 1 + 0.335 549 469 182 824 284 16;
  • 76) 0.335 549 469 182 824 284 16 × 2 = 0 + 0.671 098 938 365 648 568 32;
  • 77) 0.671 098 938 365 648 568 32 × 2 = 1 + 0.342 197 876 731 297 136 64;
  • 78) 0.342 197 876 731 297 136 64 × 2 = 0 + 0.684 395 753 462 594 273 28;
  • 79) 0.684 395 753 462 594 273 28 × 2 = 1 + 0.368 791 506 925 188 546 56;
  • 80) 0.368 791 506 925 188 546 56 × 2 = 0 + 0.737 583 013 850 377 093 12;
  • 81) 0.737 583 013 850 377 093 12 × 2 = 1 + 0.475 166 027 700 754 186 24;
  • 82) 0.475 166 027 700 754 186 24 × 2 = 0 + 0.950 332 055 401 508 372 48;

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...)


5. 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:


0.000 000 000 000 000 002 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0111 0001 1011 0110 1010 10(2)


6. Positive number before normalization:

0.000 000 000 000 000 002 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0111 0001 1011 0110 1010 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 59 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 002 12(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0111 0001 1011 0110 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0111 0001 1011 0110 1010 10(2) × 20 =

1.0011 1000 1101 1011 0101 010(2) × 2-59


8. 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 1 (a negative number)

Exponent (unadjusted): -59

Mantissa (not normalized):
1.0011 1000 1101 1011 0101 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-59 + 2(8-1) - 1 =

(-59 + 127)(10) =

68(10)


10. 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;
  • 68 ÷ 2 = 34 + 0;
  • 34 ÷ 2 = 17 + 0;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

68(10) =

0100 0100(2)


12. 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. 001 1100 0110 1101 1010 1010 =

001 1100 0110 1101 1010 1010


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

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

Exponent (8 bits) =
0100 0100

Mantissa (23 bits) =
001 1100 0110 1101 1010 1010


The base ten decimal number -0.000 000 000 000 000 002 12 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0100 0100 - 001 1100 0110 1101 1010 1010

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