-0.000 000 000 742 147 676 646 731 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 731 4(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
-0.000 000 000 742 147 676 646 731 4(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:

|-0.000 000 000 742 147 676 646 731 4| = 0.000 000 000 742 147 676 646 731 4


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: 0.000 000 000 742 147 676 646 731 4.

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) 0.000 000 000 742 147 676 646 731 4 × 2 = 0 + 0.000 000 001 484 295 353 293 462 8;
  • 2) 0.000 000 001 484 295 353 293 462 8 × 2 = 0 + 0.000 000 002 968 590 706 586 925 6;
  • 3) 0.000 000 002 968 590 706 586 925 6 × 2 = 0 + 0.000 000 005 937 181 413 173 851 2;
  • 4) 0.000 000 005 937 181 413 173 851 2 × 2 = 0 + 0.000 000 011 874 362 826 347 702 4;
  • 5) 0.000 000 011 874 362 826 347 702 4 × 2 = 0 + 0.000 000 023 748 725 652 695 404 8;
  • 6) 0.000 000 023 748 725 652 695 404 8 × 2 = 0 + 0.000 000 047 497 451 305 390 809 6;
  • 7) 0.000 000 047 497 451 305 390 809 6 × 2 = 0 + 0.000 000 094 994 902 610 781 619 2;
  • 8) 0.000 000 094 994 902 610 781 619 2 × 2 = 0 + 0.000 000 189 989 805 221 563 238 4;
  • 9) 0.000 000 189 989 805 221 563 238 4 × 2 = 0 + 0.000 000 379 979 610 443 126 476 8;
  • 10) 0.000 000 379 979 610 443 126 476 8 × 2 = 0 + 0.000 000 759 959 220 886 252 953 6;
  • 11) 0.000 000 759 959 220 886 252 953 6 × 2 = 0 + 0.000 001 519 918 441 772 505 907 2;
  • 12) 0.000 001 519 918 441 772 505 907 2 × 2 = 0 + 0.000 003 039 836 883 545 011 814 4;
  • 13) 0.000 003 039 836 883 545 011 814 4 × 2 = 0 + 0.000 006 079 673 767 090 023 628 8;
  • 14) 0.000 006 079 673 767 090 023 628 8 × 2 = 0 + 0.000 012 159 347 534 180 047 257 6;
  • 15) 0.000 012 159 347 534 180 047 257 6 × 2 = 0 + 0.000 024 318 695 068 360 094 515 2;
  • 16) 0.000 024 318 695 068 360 094 515 2 × 2 = 0 + 0.000 048 637 390 136 720 189 030 4;
  • 17) 0.000 048 637 390 136 720 189 030 4 × 2 = 0 + 0.000 097 274 780 273 440 378 060 8;
  • 18) 0.000 097 274 780 273 440 378 060 8 × 2 = 0 + 0.000 194 549 560 546 880 756 121 6;
  • 19) 0.000 194 549 560 546 880 756 121 6 × 2 = 0 + 0.000 389 099 121 093 761 512 243 2;
  • 20) 0.000 389 099 121 093 761 512 243 2 × 2 = 0 + 0.000 778 198 242 187 523 024 486 4;
  • 21) 0.000 778 198 242 187 523 024 486 4 × 2 = 0 + 0.001 556 396 484 375 046 048 972 8;
  • 22) 0.001 556 396 484 375 046 048 972 8 × 2 = 0 + 0.003 112 792 968 750 092 097 945 6;
  • 23) 0.003 112 792 968 750 092 097 945 6 × 2 = 0 + 0.006 225 585 937 500 184 195 891 2;
  • 24) 0.006 225 585 937 500 184 195 891 2 × 2 = 0 + 0.012 451 171 875 000 368 391 782 4;
  • 25) 0.012 451 171 875 000 368 391 782 4 × 2 = 0 + 0.024 902 343 750 000 736 783 564 8;
  • 26) 0.024 902 343 750 000 736 783 564 8 × 2 = 0 + 0.049 804 687 500 001 473 567 129 6;
  • 27) 0.049 804 687 500 001 473 567 129 6 × 2 = 0 + 0.099 609 375 000 002 947 134 259 2;
  • 28) 0.099 609 375 000 002 947 134 259 2 × 2 = 0 + 0.199 218 750 000 005 894 268 518 4;
  • 29) 0.199 218 750 000 005 894 268 518 4 × 2 = 0 + 0.398 437 500 000 011 788 537 036 8;
  • 30) 0.398 437 500 000 011 788 537 036 8 × 2 = 0 + 0.796 875 000 000 023 577 074 073 6;
  • 31) 0.796 875 000 000 023 577 074 073 6 × 2 = 1 + 0.593 750 000 000 047 154 148 147 2;
  • 32) 0.593 750 000 000 047 154 148 147 2 × 2 = 1 + 0.187 500 000 000 094 308 296 294 4;
  • 33) 0.187 500 000 000 094 308 296 294 4 × 2 = 0 + 0.375 000 000 000 188 616 592 588 8;
  • 34) 0.375 000 000 000 188 616 592 588 8 × 2 = 0 + 0.750 000 000 000 377 233 185 177 6;
  • 35) 0.750 000 000 000 377 233 185 177 6 × 2 = 1 + 0.500 000 000 000 754 466 370 355 2;
  • 36) 0.500 000 000 000 754 466 370 355 2 × 2 = 1 + 0.000 000 000 001 508 932 740 710 4;
  • 37) 0.000 000 000 001 508 932 740 710 4 × 2 = 0 + 0.000 000 000 003 017 865 481 420 8;
  • 38) 0.000 000 000 003 017 865 481 420 8 × 2 = 0 + 0.000 000 000 006 035 730 962 841 6;
  • 39) 0.000 000 000 006 035 730 962 841 6 × 2 = 0 + 0.000 000 000 012 071 461 925 683 2;
  • 40) 0.000 000 000 012 071 461 925 683 2 × 2 = 0 + 0.000 000 000 024 142 923 851 366 4;
  • 41) 0.000 000 000 024 142 923 851 366 4 × 2 = 0 + 0.000 000 000 048 285 847 702 732 8;
  • 42) 0.000 000 000 048 285 847 702 732 8 × 2 = 0 + 0.000 000 000 096 571 695 405 465 6;
  • 43) 0.000 000 000 096 571 695 405 465 6 × 2 = 0 + 0.000 000 000 193 143 390 810 931 2;
  • 44) 0.000 000 000 193 143 390 810 931 2 × 2 = 0 + 0.000 000 000 386 286 781 621 862 4;
  • 45) 0.000 000 000 386 286 781 621 862 4 × 2 = 0 + 0.000 000 000 772 573 563 243 724 8;
  • 46) 0.000 000 000 772 573 563 243 724 8 × 2 = 0 + 0.000 000 001 545 147 126 487 449 6;
  • 47) 0.000 000 001 545 147 126 487 449 6 × 2 = 0 + 0.000 000 003 090 294 252 974 899 2;
  • 48) 0.000 000 003 090 294 252 974 899 2 × 2 = 0 + 0.000 000 006 180 588 505 949 798 4;
  • 49) 0.000 000 006 180 588 505 949 798 4 × 2 = 0 + 0.000 000 012 361 177 011 899 596 8;
  • 50) 0.000 000 012 361 177 011 899 596 8 × 2 = 0 + 0.000 000 024 722 354 023 799 193 6;
  • 51) 0.000 000 024 722 354 023 799 193 6 × 2 = 0 + 0.000 000 049 444 708 047 598 387 2;
  • 52) 0.000 000 049 444 708 047 598 387 2 × 2 = 0 + 0.000 000 098 889 416 095 196 774 4;
  • 53) 0.000 000 098 889 416 095 196 774 4 × 2 = 0 + 0.000 000 197 778 832 190 393 548 8;
  • 54) 0.000 000 197 778 832 190 393 548 8 × 2 = 0 + 0.000 000 395 557 664 380 787 097 6;

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:


0.000 000 000 742 147 676 646 731 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 731 4(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:


0.000 000 000 742 147 676 646 731 4(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 -0.000 000 000 742 147 676 646 731 4 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