-0.000 000 000 742 147 677 07 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 07(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
-0.000 000 000 742 147 677 07(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:

|-0.000 000 000 742 147 677 07| = 0.000 000 000 742 147 677 07


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: 0.000 000 000 742 147 677 07.

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) 0.000 000 000 742 147 677 07 × 2 = 0 + 0.000 000 001 484 295 354 14;
  • 2) 0.000 000 001 484 295 354 14 × 2 = 0 + 0.000 000 002 968 590 708 28;
  • 3) 0.000 000 002 968 590 708 28 × 2 = 0 + 0.000 000 005 937 181 416 56;
  • 4) 0.000 000 005 937 181 416 56 × 2 = 0 + 0.000 000 011 874 362 833 12;
  • 5) 0.000 000 011 874 362 833 12 × 2 = 0 + 0.000 000 023 748 725 666 24;
  • 6) 0.000 000 023 748 725 666 24 × 2 = 0 + 0.000 000 047 497 451 332 48;
  • 7) 0.000 000 047 497 451 332 48 × 2 = 0 + 0.000 000 094 994 902 664 96;
  • 8) 0.000 000 094 994 902 664 96 × 2 = 0 + 0.000 000 189 989 805 329 92;
  • 9) 0.000 000 189 989 805 329 92 × 2 = 0 + 0.000 000 379 979 610 659 84;
  • 10) 0.000 000 379 979 610 659 84 × 2 = 0 + 0.000 000 759 959 221 319 68;
  • 11) 0.000 000 759 959 221 319 68 × 2 = 0 + 0.000 001 519 918 442 639 36;
  • 12) 0.000 001 519 918 442 639 36 × 2 = 0 + 0.000 003 039 836 885 278 72;
  • 13) 0.000 003 039 836 885 278 72 × 2 = 0 + 0.000 006 079 673 770 557 44;
  • 14) 0.000 006 079 673 770 557 44 × 2 = 0 + 0.000 012 159 347 541 114 88;
  • 15) 0.000 012 159 347 541 114 88 × 2 = 0 + 0.000 024 318 695 082 229 76;
  • 16) 0.000 024 318 695 082 229 76 × 2 = 0 + 0.000 048 637 390 164 459 52;
  • 17) 0.000 048 637 390 164 459 52 × 2 = 0 + 0.000 097 274 780 328 919 04;
  • 18) 0.000 097 274 780 328 919 04 × 2 = 0 + 0.000 194 549 560 657 838 08;
  • 19) 0.000 194 549 560 657 838 08 × 2 = 0 + 0.000 389 099 121 315 676 16;
  • 20) 0.000 389 099 121 315 676 16 × 2 = 0 + 0.000 778 198 242 631 352 32;
  • 21) 0.000 778 198 242 631 352 32 × 2 = 0 + 0.001 556 396 485 262 704 64;
  • 22) 0.001 556 396 485 262 704 64 × 2 = 0 + 0.003 112 792 970 525 409 28;
  • 23) 0.003 112 792 970 525 409 28 × 2 = 0 + 0.006 225 585 941 050 818 56;
  • 24) 0.006 225 585 941 050 818 56 × 2 = 0 + 0.012 451 171 882 101 637 12;
  • 25) 0.012 451 171 882 101 637 12 × 2 = 0 + 0.024 902 343 764 203 274 24;
  • 26) 0.024 902 343 764 203 274 24 × 2 = 0 + 0.049 804 687 528 406 548 48;
  • 27) 0.049 804 687 528 406 548 48 × 2 = 0 + 0.099 609 375 056 813 096 96;
  • 28) 0.099 609 375 056 813 096 96 × 2 = 0 + 0.199 218 750 113 626 193 92;
  • 29) 0.199 218 750 113 626 193 92 × 2 = 0 + 0.398 437 500 227 252 387 84;
  • 30) 0.398 437 500 227 252 387 84 × 2 = 0 + 0.796 875 000 454 504 775 68;
  • 31) 0.796 875 000 454 504 775 68 × 2 = 1 + 0.593 750 000 909 009 551 36;
  • 32) 0.593 750 000 909 009 551 36 × 2 = 1 + 0.187 500 001 818 019 102 72;
  • 33) 0.187 500 001 818 019 102 72 × 2 = 0 + 0.375 000 003 636 038 205 44;
  • 34) 0.375 000 003 636 038 205 44 × 2 = 0 + 0.750 000 007 272 076 410 88;
  • 35) 0.750 000 007 272 076 410 88 × 2 = 1 + 0.500 000 014 544 152 821 76;
  • 36) 0.500 000 014 544 152 821 76 × 2 = 1 + 0.000 000 029 088 305 643 52;
  • 37) 0.000 000 029 088 305 643 52 × 2 = 0 + 0.000 000 058 176 611 287 04;
  • 38) 0.000 000 058 176 611 287 04 × 2 = 0 + 0.000 000 116 353 222 574 08;
  • 39) 0.000 000 116 353 222 574 08 × 2 = 0 + 0.000 000 232 706 445 148 16;
  • 40) 0.000 000 232 706 445 148 16 × 2 = 0 + 0.000 000 465 412 890 296 32;
  • 41) 0.000 000 465 412 890 296 32 × 2 = 0 + 0.000 000 930 825 780 592 64;
  • 42) 0.000 000 930 825 780 592 64 × 2 = 0 + 0.000 001 861 651 561 185 28;
  • 43) 0.000 001 861 651 561 185 28 × 2 = 0 + 0.000 003 723 303 122 370 56;
  • 44) 0.000 003 723 303 122 370 56 × 2 = 0 + 0.000 007 446 606 244 741 12;
  • 45) 0.000 007 446 606 244 741 12 × 2 = 0 + 0.000 014 893 212 489 482 24;
  • 46) 0.000 014 893 212 489 482 24 × 2 = 0 + 0.000 029 786 424 978 964 48;
  • 47) 0.000 029 786 424 978 964 48 × 2 = 0 + 0.000 059 572 849 957 928 96;
  • 48) 0.000 059 572 849 957 928 96 × 2 = 0 + 0.000 119 145 699 915 857 92;
  • 49) 0.000 119 145 699 915 857 92 × 2 = 0 + 0.000 238 291 399 831 715 84;
  • 50) 0.000 238 291 399 831 715 84 × 2 = 0 + 0.000 476 582 799 663 431 68;
  • 51) 0.000 476 582 799 663 431 68 × 2 = 0 + 0.000 953 165 599 326 863 36;
  • 52) 0.000 953 165 599 326 863 36 × 2 = 0 + 0.001 906 331 198 653 726 72;
  • 53) 0.001 906 331 198 653 726 72 × 2 = 0 + 0.003 812 662 397 307 453 44;
  • 54) 0.003 812 662 397 307 453 44 × 2 = 0 + 0.007 625 324 794 614 906 88;

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:


0.000 000 000 742 147 677 07(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 677 07(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:


0.000 000 000 742 147 677 07(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 -0.000 000 000 742 147 677 07 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