-0.000 000 000 742 147 676 646 729 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 729 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
-0.000 000 000 742 147 676 646 729 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:

|-0.000 000 000 742 147 676 646 729 1| = 0.000 000 000 742 147 676 646 729 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: 0.000 000 000 742 147 676 646 729 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) 0.000 000 000 742 147 676 646 729 1 × 2 = 0 + 0.000 000 001 484 295 353 293 458 2;
  • 2) 0.000 000 001 484 295 353 293 458 2 × 2 = 0 + 0.000 000 002 968 590 706 586 916 4;
  • 3) 0.000 000 002 968 590 706 586 916 4 × 2 = 0 + 0.000 000 005 937 181 413 173 832 8;
  • 4) 0.000 000 005 937 181 413 173 832 8 × 2 = 0 + 0.000 000 011 874 362 826 347 665 6;
  • 5) 0.000 000 011 874 362 826 347 665 6 × 2 = 0 + 0.000 000 023 748 725 652 695 331 2;
  • 6) 0.000 000 023 748 725 652 695 331 2 × 2 = 0 + 0.000 000 047 497 451 305 390 662 4;
  • 7) 0.000 000 047 497 451 305 390 662 4 × 2 = 0 + 0.000 000 094 994 902 610 781 324 8;
  • 8) 0.000 000 094 994 902 610 781 324 8 × 2 = 0 + 0.000 000 189 989 805 221 562 649 6;
  • 9) 0.000 000 189 989 805 221 562 649 6 × 2 = 0 + 0.000 000 379 979 610 443 125 299 2;
  • 10) 0.000 000 379 979 610 443 125 299 2 × 2 = 0 + 0.000 000 759 959 220 886 250 598 4;
  • 11) 0.000 000 759 959 220 886 250 598 4 × 2 = 0 + 0.000 001 519 918 441 772 501 196 8;
  • 12) 0.000 001 519 918 441 772 501 196 8 × 2 = 0 + 0.000 003 039 836 883 545 002 393 6;
  • 13) 0.000 003 039 836 883 545 002 393 6 × 2 = 0 + 0.000 006 079 673 767 090 004 787 2;
  • 14) 0.000 006 079 673 767 090 004 787 2 × 2 = 0 + 0.000 012 159 347 534 180 009 574 4;
  • 15) 0.000 012 159 347 534 180 009 574 4 × 2 = 0 + 0.000 024 318 695 068 360 019 148 8;
  • 16) 0.000 024 318 695 068 360 019 148 8 × 2 = 0 + 0.000 048 637 390 136 720 038 297 6;
  • 17) 0.000 048 637 390 136 720 038 297 6 × 2 = 0 + 0.000 097 274 780 273 440 076 595 2;
  • 18) 0.000 097 274 780 273 440 076 595 2 × 2 = 0 + 0.000 194 549 560 546 880 153 190 4;
  • 19) 0.000 194 549 560 546 880 153 190 4 × 2 = 0 + 0.000 389 099 121 093 760 306 380 8;
  • 20) 0.000 389 099 121 093 760 306 380 8 × 2 = 0 + 0.000 778 198 242 187 520 612 761 6;
  • 21) 0.000 778 198 242 187 520 612 761 6 × 2 = 0 + 0.001 556 396 484 375 041 225 523 2;
  • 22) 0.001 556 396 484 375 041 225 523 2 × 2 = 0 + 0.003 112 792 968 750 082 451 046 4;
  • 23) 0.003 112 792 968 750 082 451 046 4 × 2 = 0 + 0.006 225 585 937 500 164 902 092 8;
  • 24) 0.006 225 585 937 500 164 902 092 8 × 2 = 0 + 0.012 451 171 875 000 329 804 185 6;
  • 25) 0.012 451 171 875 000 329 804 185 6 × 2 = 0 + 0.024 902 343 750 000 659 608 371 2;
  • 26) 0.024 902 343 750 000 659 608 371 2 × 2 = 0 + 0.049 804 687 500 001 319 216 742 4;
  • 27) 0.049 804 687 500 001 319 216 742 4 × 2 = 0 + 0.099 609 375 000 002 638 433 484 8;
  • 28) 0.099 609 375 000 002 638 433 484 8 × 2 = 0 + 0.199 218 750 000 005 276 866 969 6;
  • 29) 0.199 218 750 000 005 276 866 969 6 × 2 = 0 + 0.398 437 500 000 010 553 733 939 2;
  • 30) 0.398 437 500 000 010 553 733 939 2 × 2 = 0 + 0.796 875 000 000 021 107 467 878 4;
  • 31) 0.796 875 000 000 021 107 467 878 4 × 2 = 1 + 0.593 750 000 000 042 214 935 756 8;
  • 32) 0.593 750 000 000 042 214 935 756 8 × 2 = 1 + 0.187 500 000 000 084 429 871 513 6;
  • 33) 0.187 500 000 000 084 429 871 513 6 × 2 = 0 + 0.375 000 000 000 168 859 743 027 2;
  • 34) 0.375 000 000 000 168 859 743 027 2 × 2 = 0 + 0.750 000 000 000 337 719 486 054 4;
  • 35) 0.750 000 000 000 337 719 486 054 4 × 2 = 1 + 0.500 000 000 000 675 438 972 108 8;
  • 36) 0.500 000 000 000 675 438 972 108 8 × 2 = 1 + 0.000 000 000 001 350 877 944 217 6;
  • 37) 0.000 000 000 001 350 877 944 217 6 × 2 = 0 + 0.000 000 000 002 701 755 888 435 2;
  • 38) 0.000 000 000 002 701 755 888 435 2 × 2 = 0 + 0.000 000 000 005 403 511 776 870 4;
  • 39) 0.000 000 000 005 403 511 776 870 4 × 2 = 0 + 0.000 000 000 010 807 023 553 740 8;
  • 40) 0.000 000 000 010 807 023 553 740 8 × 2 = 0 + 0.000 000 000 021 614 047 107 481 6;
  • 41) 0.000 000 000 021 614 047 107 481 6 × 2 = 0 + 0.000 000 000 043 228 094 214 963 2;
  • 42) 0.000 000 000 043 228 094 214 963 2 × 2 = 0 + 0.000 000 000 086 456 188 429 926 4;
  • 43) 0.000 000 000 086 456 188 429 926 4 × 2 = 0 + 0.000 000 000 172 912 376 859 852 8;
  • 44) 0.000 000 000 172 912 376 859 852 8 × 2 = 0 + 0.000 000 000 345 824 753 719 705 6;
  • 45) 0.000 000 000 345 824 753 719 705 6 × 2 = 0 + 0.000 000 000 691 649 507 439 411 2;
  • 46) 0.000 000 000 691 649 507 439 411 2 × 2 = 0 + 0.000 000 001 383 299 014 878 822 4;
  • 47) 0.000 000 001 383 299 014 878 822 4 × 2 = 0 + 0.000 000 002 766 598 029 757 644 8;
  • 48) 0.000 000 002 766 598 029 757 644 8 × 2 = 0 + 0.000 000 005 533 196 059 515 289 6;
  • 49) 0.000 000 005 533 196 059 515 289 6 × 2 = 0 + 0.000 000 011 066 392 119 030 579 2;
  • 50) 0.000 000 011 066 392 119 030 579 2 × 2 = 0 + 0.000 000 022 132 784 238 061 158 4;
  • 51) 0.000 000 022 132 784 238 061 158 4 × 2 = 0 + 0.000 000 044 265 568 476 122 316 8;
  • 52) 0.000 000 044 265 568 476 122 316 8 × 2 = 0 + 0.000 000 088 531 136 952 244 633 6;
  • 53) 0.000 000 088 531 136 952 244 633 6 × 2 = 0 + 0.000 000 177 062 273 904 489 267 2;
  • 54) 0.000 000 177 062 273 904 489 267 2 × 2 = 0 + 0.000 000 354 124 547 808 978 534 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:


0.000 000 000 742 147 676 646 729 1(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 729 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:


0.000 000 000 742 147 676 646 729 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 -0.000 000 000 742 147 676 646 729 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