-0.000 000 000 742 147 676 646 727 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 727 3(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 646 727 3(10) to 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:

|-0.000 000 000 742 147 676 646 727 3| = 0.000 000 000 742 147 676 646 727 3


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: 0.000 000 000 742 147 676 646 727 3.

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) 0.000 000 000 742 147 676 646 727 3 × 2 = 0 + 0.000 000 001 484 295 353 293 454 6;
  • 2) 0.000 000 001 484 295 353 293 454 6 × 2 = 0 + 0.000 000 002 968 590 706 586 909 2;
  • 3) 0.000 000 002 968 590 706 586 909 2 × 2 = 0 + 0.000 000 005 937 181 413 173 818 4;
  • 4) 0.000 000 005 937 181 413 173 818 4 × 2 = 0 + 0.000 000 011 874 362 826 347 636 8;
  • 5) 0.000 000 011 874 362 826 347 636 8 × 2 = 0 + 0.000 000 023 748 725 652 695 273 6;
  • 6) 0.000 000 023 748 725 652 695 273 6 × 2 = 0 + 0.000 000 047 497 451 305 390 547 2;
  • 7) 0.000 000 047 497 451 305 390 547 2 × 2 = 0 + 0.000 000 094 994 902 610 781 094 4;
  • 8) 0.000 000 094 994 902 610 781 094 4 × 2 = 0 + 0.000 000 189 989 805 221 562 188 8;
  • 9) 0.000 000 189 989 805 221 562 188 8 × 2 = 0 + 0.000 000 379 979 610 443 124 377 6;
  • 10) 0.000 000 379 979 610 443 124 377 6 × 2 = 0 + 0.000 000 759 959 220 886 248 755 2;
  • 11) 0.000 000 759 959 220 886 248 755 2 × 2 = 0 + 0.000 001 519 918 441 772 497 510 4;
  • 12) 0.000 001 519 918 441 772 497 510 4 × 2 = 0 + 0.000 003 039 836 883 544 995 020 8;
  • 13) 0.000 003 039 836 883 544 995 020 8 × 2 = 0 + 0.000 006 079 673 767 089 990 041 6;
  • 14) 0.000 006 079 673 767 089 990 041 6 × 2 = 0 + 0.000 012 159 347 534 179 980 083 2;
  • 15) 0.000 012 159 347 534 179 980 083 2 × 2 = 0 + 0.000 024 318 695 068 359 960 166 4;
  • 16) 0.000 024 318 695 068 359 960 166 4 × 2 = 0 + 0.000 048 637 390 136 719 920 332 8;
  • 17) 0.000 048 637 390 136 719 920 332 8 × 2 = 0 + 0.000 097 274 780 273 439 840 665 6;
  • 18) 0.000 097 274 780 273 439 840 665 6 × 2 = 0 + 0.000 194 549 560 546 879 681 331 2;
  • 19) 0.000 194 549 560 546 879 681 331 2 × 2 = 0 + 0.000 389 099 121 093 759 362 662 4;
  • 20) 0.000 389 099 121 093 759 362 662 4 × 2 = 0 + 0.000 778 198 242 187 518 725 324 8;
  • 21) 0.000 778 198 242 187 518 725 324 8 × 2 = 0 + 0.001 556 396 484 375 037 450 649 6;
  • 22) 0.001 556 396 484 375 037 450 649 6 × 2 = 0 + 0.003 112 792 968 750 074 901 299 2;
  • 23) 0.003 112 792 968 750 074 901 299 2 × 2 = 0 + 0.006 225 585 937 500 149 802 598 4;
  • 24) 0.006 225 585 937 500 149 802 598 4 × 2 = 0 + 0.012 451 171 875 000 299 605 196 8;
  • 25) 0.012 451 171 875 000 299 605 196 8 × 2 = 0 + 0.024 902 343 750 000 599 210 393 6;
  • 26) 0.024 902 343 750 000 599 210 393 6 × 2 = 0 + 0.049 804 687 500 001 198 420 787 2;
  • 27) 0.049 804 687 500 001 198 420 787 2 × 2 = 0 + 0.099 609 375 000 002 396 841 574 4;
  • 28) 0.099 609 375 000 002 396 841 574 4 × 2 = 0 + 0.199 218 750 000 004 793 683 148 8;
  • 29) 0.199 218 750 000 004 793 683 148 8 × 2 = 0 + 0.398 437 500 000 009 587 366 297 6;
  • 30) 0.398 437 500 000 009 587 366 297 6 × 2 = 0 + 0.796 875 000 000 019 174 732 595 2;
  • 31) 0.796 875 000 000 019 174 732 595 2 × 2 = 1 + 0.593 750 000 000 038 349 465 190 4;
  • 32) 0.593 750 000 000 038 349 465 190 4 × 2 = 1 + 0.187 500 000 000 076 698 930 380 8;
  • 33) 0.187 500 000 000 076 698 930 380 8 × 2 = 0 + 0.375 000 000 000 153 397 860 761 6;
  • 34) 0.375 000 000 000 153 397 860 761 6 × 2 = 0 + 0.750 000 000 000 306 795 721 523 2;
  • 35) 0.750 000 000 000 306 795 721 523 2 × 2 = 1 + 0.500 000 000 000 613 591 443 046 4;
  • 36) 0.500 000 000 000 613 591 443 046 4 × 2 = 1 + 0.000 000 000 001 227 182 886 092 8;
  • 37) 0.000 000 000 001 227 182 886 092 8 × 2 = 0 + 0.000 000 000 002 454 365 772 185 6;
  • 38) 0.000 000 000 002 454 365 772 185 6 × 2 = 0 + 0.000 000 000 004 908 731 544 371 2;
  • 39) 0.000 000 000 004 908 731 544 371 2 × 2 = 0 + 0.000 000 000 009 817 463 088 742 4;
  • 40) 0.000 000 000 009 817 463 088 742 4 × 2 = 0 + 0.000 000 000 019 634 926 177 484 8;
  • 41) 0.000 000 000 019 634 926 177 484 8 × 2 = 0 + 0.000 000 000 039 269 852 354 969 6;
  • 42) 0.000 000 000 039 269 852 354 969 6 × 2 = 0 + 0.000 000 000 078 539 704 709 939 2;
  • 43) 0.000 000 000 078 539 704 709 939 2 × 2 = 0 + 0.000 000 000 157 079 409 419 878 4;
  • 44) 0.000 000 000 157 079 409 419 878 4 × 2 = 0 + 0.000 000 000 314 158 818 839 756 8;
  • 45) 0.000 000 000 314 158 818 839 756 8 × 2 = 0 + 0.000 000 000 628 317 637 679 513 6;
  • 46) 0.000 000 000 628 317 637 679 513 6 × 2 = 0 + 0.000 000 001 256 635 275 359 027 2;
  • 47) 0.000 000 001 256 635 275 359 027 2 × 2 = 0 + 0.000 000 002 513 270 550 718 054 4;
  • 48) 0.000 000 002 513 270 550 718 054 4 × 2 = 0 + 0.000 000 005 026 541 101 436 108 8;
  • 49) 0.000 000 005 026 541 101 436 108 8 × 2 = 0 + 0.000 000 010 053 082 202 872 217 6;
  • 50) 0.000 000 010 053 082 202 872 217 6 × 2 = 0 + 0.000 000 020 106 164 405 744 435 2;
  • 51) 0.000 000 020 106 164 405 744 435 2 × 2 = 0 + 0.000 000 040 212 328 811 488 870 4;
  • 52) 0.000 000 040 212 328 811 488 870 4 × 2 = 0 + 0.000 000 080 424 657 622 977 740 8;
  • 53) 0.000 000 080 424 657 622 977 740 8 × 2 = 0 + 0.000 000 160 849 315 245 955 481 6;
  • 54) 0.000 000 160 849 315 245 955 481 6 × 2 = 0 + 0.000 000 321 698 630 491 910 963 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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.000 000 000 742 147 676 646 727 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 727 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 727 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 1000 0000 0000 0000 000


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

96(10) =

0110 0000(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. 100 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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) =
0110 0000

Mantissa (23 bits) =
100 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 727 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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