-0.000 000 000 742 147 689 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 689 1(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
-0.000 000 000 742 147 689 1(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:

|-0.000 000 000 742 147 689 1| = 0.000 000 000 742 147 689 1


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: 0.000 000 000 742 147 689 1.

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) 0.000 000 000 742 147 689 1 × 2 = 0 + 0.000 000 001 484 295 378 2;
  • 2) 0.000 000 001 484 295 378 2 × 2 = 0 + 0.000 000 002 968 590 756 4;
  • 3) 0.000 000 002 968 590 756 4 × 2 = 0 + 0.000 000 005 937 181 512 8;
  • 4) 0.000 000 005 937 181 512 8 × 2 = 0 + 0.000 000 011 874 363 025 6;
  • 5) 0.000 000 011 874 363 025 6 × 2 = 0 + 0.000 000 023 748 726 051 2;
  • 6) 0.000 000 023 748 726 051 2 × 2 = 0 + 0.000 000 047 497 452 102 4;
  • 7) 0.000 000 047 497 452 102 4 × 2 = 0 + 0.000 000 094 994 904 204 8;
  • 8) 0.000 000 094 994 904 204 8 × 2 = 0 + 0.000 000 189 989 808 409 6;
  • 9) 0.000 000 189 989 808 409 6 × 2 = 0 + 0.000 000 379 979 616 819 2;
  • 10) 0.000 000 379 979 616 819 2 × 2 = 0 + 0.000 000 759 959 233 638 4;
  • 11) 0.000 000 759 959 233 638 4 × 2 = 0 + 0.000 001 519 918 467 276 8;
  • 12) 0.000 001 519 918 467 276 8 × 2 = 0 + 0.000 003 039 836 934 553 6;
  • 13) 0.000 003 039 836 934 553 6 × 2 = 0 + 0.000 006 079 673 869 107 2;
  • 14) 0.000 006 079 673 869 107 2 × 2 = 0 + 0.000 012 159 347 738 214 4;
  • 15) 0.000 012 159 347 738 214 4 × 2 = 0 + 0.000 024 318 695 476 428 8;
  • 16) 0.000 024 318 695 476 428 8 × 2 = 0 + 0.000 048 637 390 952 857 6;
  • 17) 0.000 048 637 390 952 857 6 × 2 = 0 + 0.000 097 274 781 905 715 2;
  • 18) 0.000 097 274 781 905 715 2 × 2 = 0 + 0.000 194 549 563 811 430 4;
  • 19) 0.000 194 549 563 811 430 4 × 2 = 0 + 0.000 389 099 127 622 860 8;
  • 20) 0.000 389 099 127 622 860 8 × 2 = 0 + 0.000 778 198 255 245 721 6;
  • 21) 0.000 778 198 255 245 721 6 × 2 = 0 + 0.001 556 396 510 491 443 2;
  • 22) 0.001 556 396 510 491 443 2 × 2 = 0 + 0.003 112 793 020 982 886 4;
  • 23) 0.003 112 793 020 982 886 4 × 2 = 0 + 0.006 225 586 041 965 772 8;
  • 24) 0.006 225 586 041 965 772 8 × 2 = 0 + 0.012 451 172 083 931 545 6;
  • 25) 0.012 451 172 083 931 545 6 × 2 = 0 + 0.024 902 344 167 863 091 2;
  • 26) 0.024 902 344 167 863 091 2 × 2 = 0 + 0.049 804 688 335 726 182 4;
  • 27) 0.049 804 688 335 726 182 4 × 2 = 0 + 0.099 609 376 671 452 364 8;
  • 28) 0.099 609 376 671 452 364 8 × 2 = 0 + 0.199 218 753 342 904 729 6;
  • 29) 0.199 218 753 342 904 729 6 × 2 = 0 + 0.398 437 506 685 809 459 2;
  • 30) 0.398 437 506 685 809 459 2 × 2 = 0 + 0.796 875 013 371 618 918 4;
  • 31) 0.796 875 013 371 618 918 4 × 2 = 1 + 0.593 750 026 743 237 836 8;
  • 32) 0.593 750 026 743 237 836 8 × 2 = 1 + 0.187 500 053 486 475 673 6;
  • 33) 0.187 500 053 486 475 673 6 × 2 = 0 + 0.375 000 106 972 951 347 2;
  • 34) 0.375 000 106 972 951 347 2 × 2 = 0 + 0.750 000 213 945 902 694 4;
  • 35) 0.750 000 213 945 902 694 4 × 2 = 1 + 0.500 000 427 891 805 388 8;
  • 36) 0.500 000 427 891 805 388 8 × 2 = 1 + 0.000 000 855 783 610 777 6;
  • 37) 0.000 000 855 783 610 777 6 × 2 = 0 + 0.000 001 711 567 221 555 2;
  • 38) 0.000 001 711 567 221 555 2 × 2 = 0 + 0.000 003 423 134 443 110 4;
  • 39) 0.000 003 423 134 443 110 4 × 2 = 0 + 0.000 006 846 268 886 220 8;
  • 40) 0.000 006 846 268 886 220 8 × 2 = 0 + 0.000 013 692 537 772 441 6;
  • 41) 0.000 013 692 537 772 441 6 × 2 = 0 + 0.000 027 385 075 544 883 2;
  • 42) 0.000 027 385 075 544 883 2 × 2 = 0 + 0.000 054 770 151 089 766 4;
  • 43) 0.000 054 770 151 089 766 4 × 2 = 0 + 0.000 109 540 302 179 532 8;
  • 44) 0.000 109 540 302 179 532 8 × 2 = 0 + 0.000 219 080 604 359 065 6;
  • 45) 0.000 219 080 604 359 065 6 × 2 = 0 + 0.000 438 161 208 718 131 2;
  • 46) 0.000 438 161 208 718 131 2 × 2 = 0 + 0.000 876 322 417 436 262 4;
  • 47) 0.000 876 322 417 436 262 4 × 2 = 0 + 0.001 752 644 834 872 524 8;
  • 48) 0.001 752 644 834 872 524 8 × 2 = 0 + 0.003 505 289 669 745 049 6;
  • 49) 0.003 505 289 669 745 049 6 × 2 = 0 + 0.007 010 579 339 490 099 2;
  • 50) 0.007 010 579 339 490 099 2 × 2 = 0 + 0.014 021 158 678 980 198 4;
  • 51) 0.014 021 158 678 980 198 4 × 2 = 0 + 0.028 042 317 357 960 396 8;
  • 52) 0.028 042 317 357 960 396 8 × 2 = 0 + 0.056 084 634 715 920 793 6;
  • 53) 0.056 084 634 715 920 793 6 × 2 = 0 + 0.112 169 269 431 841 587 2;
  • 54) 0.112 169 269 431 841 587 2 × 2 = 0 + 0.224 338 538 863 683 174 4;

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:


0.000 000 000 742 147 689 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 689 1(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:


0.000 000 000 742 147 689 1(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 -0.000 000 000 742 147 689 1 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