-0.000 000 000 742 147 687 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 687 8(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
-0.000 000 000 742 147 687 8(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:

|-0.000 000 000 742 147 687 8| = 0.000 000 000 742 147 687 8


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: 0.000 000 000 742 147 687 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) 0.000 000 000 742 147 687 8 × 2 = 0 + 0.000 000 001 484 295 375 6;
  • 2) 0.000 000 001 484 295 375 6 × 2 = 0 + 0.000 000 002 968 590 751 2;
  • 3) 0.000 000 002 968 590 751 2 × 2 = 0 + 0.000 000 005 937 181 502 4;
  • 4) 0.000 000 005 937 181 502 4 × 2 = 0 + 0.000 000 011 874 363 004 8;
  • 5) 0.000 000 011 874 363 004 8 × 2 = 0 + 0.000 000 023 748 726 009 6;
  • 6) 0.000 000 023 748 726 009 6 × 2 = 0 + 0.000 000 047 497 452 019 2;
  • 7) 0.000 000 047 497 452 019 2 × 2 = 0 + 0.000 000 094 994 904 038 4;
  • 8) 0.000 000 094 994 904 038 4 × 2 = 0 + 0.000 000 189 989 808 076 8;
  • 9) 0.000 000 189 989 808 076 8 × 2 = 0 + 0.000 000 379 979 616 153 6;
  • 10) 0.000 000 379 979 616 153 6 × 2 = 0 + 0.000 000 759 959 232 307 2;
  • 11) 0.000 000 759 959 232 307 2 × 2 = 0 + 0.000 001 519 918 464 614 4;
  • 12) 0.000 001 519 918 464 614 4 × 2 = 0 + 0.000 003 039 836 929 228 8;
  • 13) 0.000 003 039 836 929 228 8 × 2 = 0 + 0.000 006 079 673 858 457 6;
  • 14) 0.000 006 079 673 858 457 6 × 2 = 0 + 0.000 012 159 347 716 915 2;
  • 15) 0.000 012 159 347 716 915 2 × 2 = 0 + 0.000 024 318 695 433 830 4;
  • 16) 0.000 024 318 695 433 830 4 × 2 = 0 + 0.000 048 637 390 867 660 8;
  • 17) 0.000 048 637 390 867 660 8 × 2 = 0 + 0.000 097 274 781 735 321 6;
  • 18) 0.000 097 274 781 735 321 6 × 2 = 0 + 0.000 194 549 563 470 643 2;
  • 19) 0.000 194 549 563 470 643 2 × 2 = 0 + 0.000 389 099 126 941 286 4;
  • 20) 0.000 389 099 126 941 286 4 × 2 = 0 + 0.000 778 198 253 882 572 8;
  • 21) 0.000 778 198 253 882 572 8 × 2 = 0 + 0.001 556 396 507 765 145 6;
  • 22) 0.001 556 396 507 765 145 6 × 2 = 0 + 0.003 112 793 015 530 291 2;
  • 23) 0.003 112 793 015 530 291 2 × 2 = 0 + 0.006 225 586 031 060 582 4;
  • 24) 0.006 225 586 031 060 582 4 × 2 = 0 + 0.012 451 172 062 121 164 8;
  • 25) 0.012 451 172 062 121 164 8 × 2 = 0 + 0.024 902 344 124 242 329 6;
  • 26) 0.024 902 344 124 242 329 6 × 2 = 0 + 0.049 804 688 248 484 659 2;
  • 27) 0.049 804 688 248 484 659 2 × 2 = 0 + 0.099 609 376 496 969 318 4;
  • 28) 0.099 609 376 496 969 318 4 × 2 = 0 + 0.199 218 752 993 938 636 8;
  • 29) 0.199 218 752 993 938 636 8 × 2 = 0 + 0.398 437 505 987 877 273 6;
  • 30) 0.398 437 505 987 877 273 6 × 2 = 0 + 0.796 875 011 975 754 547 2;
  • 31) 0.796 875 011 975 754 547 2 × 2 = 1 + 0.593 750 023 951 509 094 4;
  • 32) 0.593 750 023 951 509 094 4 × 2 = 1 + 0.187 500 047 903 018 188 8;
  • 33) 0.187 500 047 903 018 188 8 × 2 = 0 + 0.375 000 095 806 036 377 6;
  • 34) 0.375 000 095 806 036 377 6 × 2 = 0 + 0.750 000 191 612 072 755 2;
  • 35) 0.750 000 191 612 072 755 2 × 2 = 1 + 0.500 000 383 224 145 510 4;
  • 36) 0.500 000 383 224 145 510 4 × 2 = 1 + 0.000 000 766 448 291 020 8;
  • 37) 0.000 000 766 448 291 020 8 × 2 = 0 + 0.000 001 532 896 582 041 6;
  • 38) 0.000 001 532 896 582 041 6 × 2 = 0 + 0.000 003 065 793 164 083 2;
  • 39) 0.000 003 065 793 164 083 2 × 2 = 0 + 0.000 006 131 586 328 166 4;
  • 40) 0.000 006 131 586 328 166 4 × 2 = 0 + 0.000 012 263 172 656 332 8;
  • 41) 0.000 012 263 172 656 332 8 × 2 = 0 + 0.000 024 526 345 312 665 6;
  • 42) 0.000 024 526 345 312 665 6 × 2 = 0 + 0.000 049 052 690 625 331 2;
  • 43) 0.000 049 052 690 625 331 2 × 2 = 0 + 0.000 098 105 381 250 662 4;
  • 44) 0.000 098 105 381 250 662 4 × 2 = 0 + 0.000 196 210 762 501 324 8;
  • 45) 0.000 196 210 762 501 324 8 × 2 = 0 + 0.000 392 421 525 002 649 6;
  • 46) 0.000 392 421 525 002 649 6 × 2 = 0 + 0.000 784 843 050 005 299 2;
  • 47) 0.000 784 843 050 005 299 2 × 2 = 0 + 0.001 569 686 100 010 598 4;
  • 48) 0.001 569 686 100 010 598 4 × 2 = 0 + 0.003 139 372 200 021 196 8;
  • 49) 0.003 139 372 200 021 196 8 × 2 = 0 + 0.006 278 744 400 042 393 6;
  • 50) 0.006 278 744 400 042 393 6 × 2 = 0 + 0.012 557 488 800 084 787 2;
  • 51) 0.012 557 488 800 084 787 2 × 2 = 0 + 0.025 114 977 600 169 574 4;
  • 52) 0.025 114 977 600 169 574 4 × 2 = 0 + 0.050 229 955 200 339 148 8;
  • 53) 0.050 229 955 200 339 148 8 × 2 = 0 + 0.100 459 910 400 678 297 6;
  • 54) 0.100 459 910 400 678 297 6 × 2 = 0 + 0.200 919 820 801 356 595 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:


0.000 000 000 742 147 687 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 687 8(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:


0.000 000 000 742 147 687 8(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 -0.000 000 000 742 147 687 8 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