-0.000 000 000 742 147 676 647 79 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 647 79(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
-0.000 000 000 742 147 676 647 79(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:

|-0.000 000 000 742 147 676 647 79| = 0.000 000 000 742 147 676 647 79


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: 0.000 000 000 742 147 676 647 79.

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) 0.000 000 000 742 147 676 647 79 × 2 = 0 + 0.000 000 001 484 295 353 295 58;
  • 2) 0.000 000 001 484 295 353 295 58 × 2 = 0 + 0.000 000 002 968 590 706 591 16;
  • 3) 0.000 000 002 968 590 706 591 16 × 2 = 0 + 0.000 000 005 937 181 413 182 32;
  • 4) 0.000 000 005 937 181 413 182 32 × 2 = 0 + 0.000 000 011 874 362 826 364 64;
  • 5) 0.000 000 011 874 362 826 364 64 × 2 = 0 + 0.000 000 023 748 725 652 729 28;
  • 6) 0.000 000 023 748 725 652 729 28 × 2 = 0 + 0.000 000 047 497 451 305 458 56;
  • 7) 0.000 000 047 497 451 305 458 56 × 2 = 0 + 0.000 000 094 994 902 610 917 12;
  • 8) 0.000 000 094 994 902 610 917 12 × 2 = 0 + 0.000 000 189 989 805 221 834 24;
  • 9) 0.000 000 189 989 805 221 834 24 × 2 = 0 + 0.000 000 379 979 610 443 668 48;
  • 10) 0.000 000 379 979 610 443 668 48 × 2 = 0 + 0.000 000 759 959 220 887 336 96;
  • 11) 0.000 000 759 959 220 887 336 96 × 2 = 0 + 0.000 001 519 918 441 774 673 92;
  • 12) 0.000 001 519 918 441 774 673 92 × 2 = 0 + 0.000 003 039 836 883 549 347 84;
  • 13) 0.000 003 039 836 883 549 347 84 × 2 = 0 + 0.000 006 079 673 767 098 695 68;
  • 14) 0.000 006 079 673 767 098 695 68 × 2 = 0 + 0.000 012 159 347 534 197 391 36;
  • 15) 0.000 012 159 347 534 197 391 36 × 2 = 0 + 0.000 024 318 695 068 394 782 72;
  • 16) 0.000 024 318 695 068 394 782 72 × 2 = 0 + 0.000 048 637 390 136 789 565 44;
  • 17) 0.000 048 637 390 136 789 565 44 × 2 = 0 + 0.000 097 274 780 273 579 130 88;
  • 18) 0.000 097 274 780 273 579 130 88 × 2 = 0 + 0.000 194 549 560 547 158 261 76;
  • 19) 0.000 194 549 560 547 158 261 76 × 2 = 0 + 0.000 389 099 121 094 316 523 52;
  • 20) 0.000 389 099 121 094 316 523 52 × 2 = 0 + 0.000 778 198 242 188 633 047 04;
  • 21) 0.000 778 198 242 188 633 047 04 × 2 = 0 + 0.001 556 396 484 377 266 094 08;
  • 22) 0.001 556 396 484 377 266 094 08 × 2 = 0 + 0.003 112 792 968 754 532 188 16;
  • 23) 0.003 112 792 968 754 532 188 16 × 2 = 0 + 0.006 225 585 937 509 064 376 32;
  • 24) 0.006 225 585 937 509 064 376 32 × 2 = 0 + 0.012 451 171 875 018 128 752 64;
  • 25) 0.012 451 171 875 018 128 752 64 × 2 = 0 + 0.024 902 343 750 036 257 505 28;
  • 26) 0.024 902 343 750 036 257 505 28 × 2 = 0 + 0.049 804 687 500 072 515 010 56;
  • 27) 0.049 804 687 500 072 515 010 56 × 2 = 0 + 0.099 609 375 000 145 030 021 12;
  • 28) 0.099 609 375 000 145 030 021 12 × 2 = 0 + 0.199 218 750 000 290 060 042 24;
  • 29) 0.199 218 750 000 290 060 042 24 × 2 = 0 + 0.398 437 500 000 580 120 084 48;
  • 30) 0.398 437 500 000 580 120 084 48 × 2 = 0 + 0.796 875 000 001 160 240 168 96;
  • 31) 0.796 875 000 001 160 240 168 96 × 2 = 1 + 0.593 750 000 002 320 480 337 92;
  • 32) 0.593 750 000 002 320 480 337 92 × 2 = 1 + 0.187 500 000 004 640 960 675 84;
  • 33) 0.187 500 000 004 640 960 675 84 × 2 = 0 + 0.375 000 000 009 281 921 351 68;
  • 34) 0.375 000 000 009 281 921 351 68 × 2 = 0 + 0.750 000 000 018 563 842 703 36;
  • 35) 0.750 000 000 018 563 842 703 36 × 2 = 1 + 0.500 000 000 037 127 685 406 72;
  • 36) 0.500 000 000 037 127 685 406 72 × 2 = 1 + 0.000 000 000 074 255 370 813 44;
  • 37) 0.000 000 000 074 255 370 813 44 × 2 = 0 + 0.000 000 000 148 510 741 626 88;
  • 38) 0.000 000 000 148 510 741 626 88 × 2 = 0 + 0.000 000 000 297 021 483 253 76;
  • 39) 0.000 000 000 297 021 483 253 76 × 2 = 0 + 0.000 000 000 594 042 966 507 52;
  • 40) 0.000 000 000 594 042 966 507 52 × 2 = 0 + 0.000 000 001 188 085 933 015 04;
  • 41) 0.000 000 001 188 085 933 015 04 × 2 = 0 + 0.000 000 002 376 171 866 030 08;
  • 42) 0.000 000 002 376 171 866 030 08 × 2 = 0 + 0.000 000 004 752 343 732 060 16;
  • 43) 0.000 000 004 752 343 732 060 16 × 2 = 0 + 0.000 000 009 504 687 464 120 32;
  • 44) 0.000 000 009 504 687 464 120 32 × 2 = 0 + 0.000 000 019 009 374 928 240 64;
  • 45) 0.000 000 019 009 374 928 240 64 × 2 = 0 + 0.000 000 038 018 749 856 481 28;
  • 46) 0.000 000 038 018 749 856 481 28 × 2 = 0 + 0.000 000 076 037 499 712 962 56;
  • 47) 0.000 000 076 037 499 712 962 56 × 2 = 0 + 0.000 000 152 074 999 425 925 12;
  • 48) 0.000 000 152 074 999 425 925 12 × 2 = 0 + 0.000 000 304 149 998 851 850 24;
  • 49) 0.000 000 304 149 998 851 850 24 × 2 = 0 + 0.000 000 608 299 997 703 700 48;
  • 50) 0.000 000 608 299 997 703 700 48 × 2 = 0 + 0.000 001 216 599 995 407 400 96;
  • 51) 0.000 001 216 599 995 407 400 96 × 2 = 0 + 0.000 002 433 199 990 814 801 92;
  • 52) 0.000 002 433 199 990 814 801 92 × 2 = 0 + 0.000 004 866 399 981 629 603 84;
  • 53) 0.000 004 866 399 981 629 603 84 × 2 = 0 + 0.000 009 732 799 963 259 207 68;
  • 54) 0.000 009 732 799 963 259 207 68 × 2 = 0 + 0.000 019 465 599 926 518 415 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:


0.000 000 000 742 147 676 647 79(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 647 79(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:


0.000 000 000 742 147 676 647 79(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 -0.000 000 000 742 147 676 647 79 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