-0.000 000 000 742 147 677 04 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 04(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 04(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 04| = 0.000 000 000 742 147 677 04


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 04.

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 04 × 2 = 0 + 0.000 000 001 484 295 354 08;
  • 2) 0.000 000 001 484 295 354 08 × 2 = 0 + 0.000 000 002 968 590 708 16;
  • 3) 0.000 000 002 968 590 708 16 × 2 = 0 + 0.000 000 005 937 181 416 32;
  • 4) 0.000 000 005 937 181 416 32 × 2 = 0 + 0.000 000 011 874 362 832 64;
  • 5) 0.000 000 011 874 362 832 64 × 2 = 0 + 0.000 000 023 748 725 665 28;
  • 6) 0.000 000 023 748 725 665 28 × 2 = 0 + 0.000 000 047 497 451 330 56;
  • 7) 0.000 000 047 497 451 330 56 × 2 = 0 + 0.000 000 094 994 902 661 12;
  • 8) 0.000 000 094 994 902 661 12 × 2 = 0 + 0.000 000 189 989 805 322 24;
  • 9) 0.000 000 189 989 805 322 24 × 2 = 0 + 0.000 000 379 979 610 644 48;
  • 10) 0.000 000 379 979 610 644 48 × 2 = 0 + 0.000 000 759 959 221 288 96;
  • 11) 0.000 000 759 959 221 288 96 × 2 = 0 + 0.000 001 519 918 442 577 92;
  • 12) 0.000 001 519 918 442 577 92 × 2 = 0 + 0.000 003 039 836 885 155 84;
  • 13) 0.000 003 039 836 885 155 84 × 2 = 0 + 0.000 006 079 673 770 311 68;
  • 14) 0.000 006 079 673 770 311 68 × 2 = 0 + 0.000 012 159 347 540 623 36;
  • 15) 0.000 012 159 347 540 623 36 × 2 = 0 + 0.000 024 318 695 081 246 72;
  • 16) 0.000 024 318 695 081 246 72 × 2 = 0 + 0.000 048 637 390 162 493 44;
  • 17) 0.000 048 637 390 162 493 44 × 2 = 0 + 0.000 097 274 780 324 986 88;
  • 18) 0.000 097 274 780 324 986 88 × 2 = 0 + 0.000 194 549 560 649 973 76;
  • 19) 0.000 194 549 560 649 973 76 × 2 = 0 + 0.000 389 099 121 299 947 52;
  • 20) 0.000 389 099 121 299 947 52 × 2 = 0 + 0.000 778 198 242 599 895 04;
  • 21) 0.000 778 198 242 599 895 04 × 2 = 0 + 0.001 556 396 485 199 790 08;
  • 22) 0.001 556 396 485 199 790 08 × 2 = 0 + 0.003 112 792 970 399 580 16;
  • 23) 0.003 112 792 970 399 580 16 × 2 = 0 + 0.006 225 585 940 799 160 32;
  • 24) 0.006 225 585 940 799 160 32 × 2 = 0 + 0.012 451 171 881 598 320 64;
  • 25) 0.012 451 171 881 598 320 64 × 2 = 0 + 0.024 902 343 763 196 641 28;
  • 26) 0.024 902 343 763 196 641 28 × 2 = 0 + 0.049 804 687 526 393 282 56;
  • 27) 0.049 804 687 526 393 282 56 × 2 = 0 + 0.099 609 375 052 786 565 12;
  • 28) 0.099 609 375 052 786 565 12 × 2 = 0 + 0.199 218 750 105 573 130 24;
  • 29) 0.199 218 750 105 573 130 24 × 2 = 0 + 0.398 437 500 211 146 260 48;
  • 30) 0.398 437 500 211 146 260 48 × 2 = 0 + 0.796 875 000 422 292 520 96;
  • 31) 0.796 875 000 422 292 520 96 × 2 = 1 + 0.593 750 000 844 585 041 92;
  • 32) 0.593 750 000 844 585 041 92 × 2 = 1 + 0.187 500 001 689 170 083 84;
  • 33) 0.187 500 001 689 170 083 84 × 2 = 0 + 0.375 000 003 378 340 167 68;
  • 34) 0.375 000 003 378 340 167 68 × 2 = 0 + 0.750 000 006 756 680 335 36;
  • 35) 0.750 000 006 756 680 335 36 × 2 = 1 + 0.500 000 013 513 360 670 72;
  • 36) 0.500 000 013 513 360 670 72 × 2 = 1 + 0.000 000 027 026 721 341 44;
  • 37) 0.000 000 027 026 721 341 44 × 2 = 0 + 0.000 000 054 053 442 682 88;
  • 38) 0.000 000 054 053 442 682 88 × 2 = 0 + 0.000 000 108 106 885 365 76;
  • 39) 0.000 000 108 106 885 365 76 × 2 = 0 + 0.000 000 216 213 770 731 52;
  • 40) 0.000 000 216 213 770 731 52 × 2 = 0 + 0.000 000 432 427 541 463 04;
  • 41) 0.000 000 432 427 541 463 04 × 2 = 0 + 0.000 000 864 855 082 926 08;
  • 42) 0.000 000 864 855 082 926 08 × 2 = 0 + 0.000 001 729 710 165 852 16;
  • 43) 0.000 001 729 710 165 852 16 × 2 = 0 + 0.000 003 459 420 331 704 32;
  • 44) 0.000 003 459 420 331 704 32 × 2 = 0 + 0.000 006 918 840 663 408 64;
  • 45) 0.000 006 918 840 663 408 64 × 2 = 0 + 0.000 013 837 681 326 817 28;
  • 46) 0.000 013 837 681 326 817 28 × 2 = 0 + 0.000 027 675 362 653 634 56;
  • 47) 0.000 027 675 362 653 634 56 × 2 = 0 + 0.000 055 350 725 307 269 12;
  • 48) 0.000 055 350 725 307 269 12 × 2 = 0 + 0.000 110 701 450 614 538 24;
  • 49) 0.000 110 701 450 614 538 24 × 2 = 0 + 0.000 221 402 901 229 076 48;
  • 50) 0.000 221 402 901 229 076 48 × 2 = 0 + 0.000 442 805 802 458 152 96;
  • 51) 0.000 442 805 802 458 152 96 × 2 = 0 + 0.000 885 611 604 916 305 92;
  • 52) 0.000 885 611 604 916 305 92 × 2 = 0 + 0.001 771 223 209 832 611 84;
  • 53) 0.001 771 223 209 832 611 84 × 2 = 0 + 0.003 542 446 419 665 223 68;
  • 54) 0.003 542 446 419 665 223 68 × 2 = 0 + 0.007 084 892 839 330 447 36;

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 04(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 04(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 04(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 04 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