-0.000 000 000 742 147 676 646 730 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 730 2(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
-0.000 000 000 742 147 676 646 730 2(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:

|-0.000 000 000 742 147 676 646 730 2| = 0.000 000 000 742 147 676 646 730 2


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: 0.000 000 000 742 147 676 646 730 2.

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) 0.000 000 000 742 147 676 646 730 2 × 2 = 0 + 0.000 000 001 484 295 353 293 460 4;
  • 2) 0.000 000 001 484 295 353 293 460 4 × 2 = 0 + 0.000 000 002 968 590 706 586 920 8;
  • 3) 0.000 000 002 968 590 706 586 920 8 × 2 = 0 + 0.000 000 005 937 181 413 173 841 6;
  • 4) 0.000 000 005 937 181 413 173 841 6 × 2 = 0 + 0.000 000 011 874 362 826 347 683 2;
  • 5) 0.000 000 011 874 362 826 347 683 2 × 2 = 0 + 0.000 000 023 748 725 652 695 366 4;
  • 6) 0.000 000 023 748 725 652 695 366 4 × 2 = 0 + 0.000 000 047 497 451 305 390 732 8;
  • 7) 0.000 000 047 497 451 305 390 732 8 × 2 = 0 + 0.000 000 094 994 902 610 781 465 6;
  • 8) 0.000 000 094 994 902 610 781 465 6 × 2 = 0 + 0.000 000 189 989 805 221 562 931 2;
  • 9) 0.000 000 189 989 805 221 562 931 2 × 2 = 0 + 0.000 000 379 979 610 443 125 862 4;
  • 10) 0.000 000 379 979 610 443 125 862 4 × 2 = 0 + 0.000 000 759 959 220 886 251 724 8;
  • 11) 0.000 000 759 959 220 886 251 724 8 × 2 = 0 + 0.000 001 519 918 441 772 503 449 6;
  • 12) 0.000 001 519 918 441 772 503 449 6 × 2 = 0 + 0.000 003 039 836 883 545 006 899 2;
  • 13) 0.000 003 039 836 883 545 006 899 2 × 2 = 0 + 0.000 006 079 673 767 090 013 798 4;
  • 14) 0.000 006 079 673 767 090 013 798 4 × 2 = 0 + 0.000 012 159 347 534 180 027 596 8;
  • 15) 0.000 012 159 347 534 180 027 596 8 × 2 = 0 + 0.000 024 318 695 068 360 055 193 6;
  • 16) 0.000 024 318 695 068 360 055 193 6 × 2 = 0 + 0.000 048 637 390 136 720 110 387 2;
  • 17) 0.000 048 637 390 136 720 110 387 2 × 2 = 0 + 0.000 097 274 780 273 440 220 774 4;
  • 18) 0.000 097 274 780 273 440 220 774 4 × 2 = 0 + 0.000 194 549 560 546 880 441 548 8;
  • 19) 0.000 194 549 560 546 880 441 548 8 × 2 = 0 + 0.000 389 099 121 093 760 883 097 6;
  • 20) 0.000 389 099 121 093 760 883 097 6 × 2 = 0 + 0.000 778 198 242 187 521 766 195 2;
  • 21) 0.000 778 198 242 187 521 766 195 2 × 2 = 0 + 0.001 556 396 484 375 043 532 390 4;
  • 22) 0.001 556 396 484 375 043 532 390 4 × 2 = 0 + 0.003 112 792 968 750 087 064 780 8;
  • 23) 0.003 112 792 968 750 087 064 780 8 × 2 = 0 + 0.006 225 585 937 500 174 129 561 6;
  • 24) 0.006 225 585 937 500 174 129 561 6 × 2 = 0 + 0.012 451 171 875 000 348 259 123 2;
  • 25) 0.012 451 171 875 000 348 259 123 2 × 2 = 0 + 0.024 902 343 750 000 696 518 246 4;
  • 26) 0.024 902 343 750 000 696 518 246 4 × 2 = 0 + 0.049 804 687 500 001 393 036 492 8;
  • 27) 0.049 804 687 500 001 393 036 492 8 × 2 = 0 + 0.099 609 375 000 002 786 072 985 6;
  • 28) 0.099 609 375 000 002 786 072 985 6 × 2 = 0 + 0.199 218 750 000 005 572 145 971 2;
  • 29) 0.199 218 750 000 005 572 145 971 2 × 2 = 0 + 0.398 437 500 000 011 144 291 942 4;
  • 30) 0.398 437 500 000 011 144 291 942 4 × 2 = 0 + 0.796 875 000 000 022 288 583 884 8;
  • 31) 0.796 875 000 000 022 288 583 884 8 × 2 = 1 + 0.593 750 000 000 044 577 167 769 6;
  • 32) 0.593 750 000 000 044 577 167 769 6 × 2 = 1 + 0.187 500 000 000 089 154 335 539 2;
  • 33) 0.187 500 000 000 089 154 335 539 2 × 2 = 0 + 0.375 000 000 000 178 308 671 078 4;
  • 34) 0.375 000 000 000 178 308 671 078 4 × 2 = 0 + 0.750 000 000 000 356 617 342 156 8;
  • 35) 0.750 000 000 000 356 617 342 156 8 × 2 = 1 + 0.500 000 000 000 713 234 684 313 6;
  • 36) 0.500 000 000 000 713 234 684 313 6 × 2 = 1 + 0.000 000 000 001 426 469 368 627 2;
  • 37) 0.000 000 000 001 426 469 368 627 2 × 2 = 0 + 0.000 000 000 002 852 938 737 254 4;
  • 38) 0.000 000 000 002 852 938 737 254 4 × 2 = 0 + 0.000 000 000 005 705 877 474 508 8;
  • 39) 0.000 000 000 005 705 877 474 508 8 × 2 = 0 + 0.000 000 000 011 411 754 949 017 6;
  • 40) 0.000 000 000 011 411 754 949 017 6 × 2 = 0 + 0.000 000 000 022 823 509 898 035 2;
  • 41) 0.000 000 000 022 823 509 898 035 2 × 2 = 0 + 0.000 000 000 045 647 019 796 070 4;
  • 42) 0.000 000 000 045 647 019 796 070 4 × 2 = 0 + 0.000 000 000 091 294 039 592 140 8;
  • 43) 0.000 000 000 091 294 039 592 140 8 × 2 = 0 + 0.000 000 000 182 588 079 184 281 6;
  • 44) 0.000 000 000 182 588 079 184 281 6 × 2 = 0 + 0.000 000 000 365 176 158 368 563 2;
  • 45) 0.000 000 000 365 176 158 368 563 2 × 2 = 0 + 0.000 000 000 730 352 316 737 126 4;
  • 46) 0.000 000 000 730 352 316 737 126 4 × 2 = 0 + 0.000 000 001 460 704 633 474 252 8;
  • 47) 0.000 000 001 460 704 633 474 252 8 × 2 = 0 + 0.000 000 002 921 409 266 948 505 6;
  • 48) 0.000 000 002 921 409 266 948 505 6 × 2 = 0 + 0.000 000 005 842 818 533 897 011 2;
  • 49) 0.000 000 005 842 818 533 897 011 2 × 2 = 0 + 0.000 000 011 685 637 067 794 022 4;
  • 50) 0.000 000 011 685 637 067 794 022 4 × 2 = 0 + 0.000 000 023 371 274 135 588 044 8;
  • 51) 0.000 000 023 371 274 135 588 044 8 × 2 = 0 + 0.000 000 046 742 548 271 176 089 6;
  • 52) 0.000 000 046 742 548 271 176 089 6 × 2 = 0 + 0.000 000 093 485 096 542 352 179 2;
  • 53) 0.000 000 093 485 096 542 352 179 2 × 2 = 0 + 0.000 000 186 970 193 084 704 358 4;
  • 54) 0.000 000 186 970 193 084 704 358 4 × 2 = 0 + 0.000 000 373 940 386 169 408 716 8;

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:


0.000 000 000 742 147 676 646 730 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 730 2(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:


0.000 000 000 742 147 676 646 730 2(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 -0.000 000 000 742 147 676 646 730 2 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