-0.000 000 000 742 147 478 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 478(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
-0.000 000 000 742 147 478(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:

|-0.000 000 000 742 147 478| = 0.000 000 000 742 147 478


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: 0.000 000 000 742 147 478.

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) 0.000 000 000 742 147 478 × 2 = 0 + 0.000 000 001 484 294 956;
  • 2) 0.000 000 001 484 294 956 × 2 = 0 + 0.000 000 002 968 589 912;
  • 3) 0.000 000 002 968 589 912 × 2 = 0 + 0.000 000 005 937 179 824;
  • 4) 0.000 000 005 937 179 824 × 2 = 0 + 0.000 000 011 874 359 648;
  • 5) 0.000 000 011 874 359 648 × 2 = 0 + 0.000 000 023 748 719 296;
  • 6) 0.000 000 023 748 719 296 × 2 = 0 + 0.000 000 047 497 438 592;
  • 7) 0.000 000 047 497 438 592 × 2 = 0 + 0.000 000 094 994 877 184;
  • 8) 0.000 000 094 994 877 184 × 2 = 0 + 0.000 000 189 989 754 368;
  • 9) 0.000 000 189 989 754 368 × 2 = 0 + 0.000 000 379 979 508 736;
  • 10) 0.000 000 379 979 508 736 × 2 = 0 + 0.000 000 759 959 017 472;
  • 11) 0.000 000 759 959 017 472 × 2 = 0 + 0.000 001 519 918 034 944;
  • 12) 0.000 001 519 918 034 944 × 2 = 0 + 0.000 003 039 836 069 888;
  • 13) 0.000 003 039 836 069 888 × 2 = 0 + 0.000 006 079 672 139 776;
  • 14) 0.000 006 079 672 139 776 × 2 = 0 + 0.000 012 159 344 279 552;
  • 15) 0.000 012 159 344 279 552 × 2 = 0 + 0.000 024 318 688 559 104;
  • 16) 0.000 024 318 688 559 104 × 2 = 0 + 0.000 048 637 377 118 208;
  • 17) 0.000 048 637 377 118 208 × 2 = 0 + 0.000 097 274 754 236 416;
  • 18) 0.000 097 274 754 236 416 × 2 = 0 + 0.000 194 549 508 472 832;
  • 19) 0.000 194 549 508 472 832 × 2 = 0 + 0.000 389 099 016 945 664;
  • 20) 0.000 389 099 016 945 664 × 2 = 0 + 0.000 778 198 033 891 328;
  • 21) 0.000 778 198 033 891 328 × 2 = 0 + 0.001 556 396 067 782 656;
  • 22) 0.001 556 396 067 782 656 × 2 = 0 + 0.003 112 792 135 565 312;
  • 23) 0.003 112 792 135 565 312 × 2 = 0 + 0.006 225 584 271 130 624;
  • 24) 0.006 225 584 271 130 624 × 2 = 0 + 0.012 451 168 542 261 248;
  • 25) 0.012 451 168 542 261 248 × 2 = 0 + 0.024 902 337 084 522 496;
  • 26) 0.024 902 337 084 522 496 × 2 = 0 + 0.049 804 674 169 044 992;
  • 27) 0.049 804 674 169 044 992 × 2 = 0 + 0.099 609 348 338 089 984;
  • 28) 0.099 609 348 338 089 984 × 2 = 0 + 0.199 218 696 676 179 968;
  • 29) 0.199 218 696 676 179 968 × 2 = 0 + 0.398 437 393 352 359 936;
  • 30) 0.398 437 393 352 359 936 × 2 = 0 + 0.796 874 786 704 719 872;
  • 31) 0.796 874 786 704 719 872 × 2 = 1 + 0.593 749 573 409 439 744;
  • 32) 0.593 749 573 409 439 744 × 2 = 1 + 0.187 499 146 818 879 488;
  • 33) 0.187 499 146 818 879 488 × 2 = 0 + 0.374 998 293 637 758 976;
  • 34) 0.374 998 293 637 758 976 × 2 = 0 + 0.749 996 587 275 517 952;
  • 35) 0.749 996 587 275 517 952 × 2 = 1 + 0.499 993 174 551 035 904;
  • 36) 0.499 993 174 551 035 904 × 2 = 0 + 0.999 986 349 102 071 808;
  • 37) 0.999 986 349 102 071 808 × 2 = 1 + 0.999 972 698 204 143 616;
  • 38) 0.999 972 698 204 143 616 × 2 = 1 + 0.999 945 396 408 287 232;
  • 39) 0.999 945 396 408 287 232 × 2 = 1 + 0.999 890 792 816 574 464;
  • 40) 0.999 890 792 816 574 464 × 2 = 1 + 0.999 781 585 633 148 928;
  • 41) 0.999 781 585 633 148 928 × 2 = 1 + 0.999 563 171 266 297 856;
  • 42) 0.999 563 171 266 297 856 × 2 = 1 + 0.999 126 342 532 595 712;
  • 43) 0.999 126 342 532 595 712 × 2 = 1 + 0.998 252 685 065 191 424;
  • 44) 0.998 252 685 065 191 424 × 2 = 1 + 0.996 505 370 130 382 848;
  • 45) 0.996 505 370 130 382 848 × 2 = 1 + 0.993 010 740 260 765 696;
  • 46) 0.993 010 740 260 765 696 × 2 = 1 + 0.986 021 480 521 531 392;
  • 47) 0.986 021 480 521 531 392 × 2 = 1 + 0.972 042 961 043 062 784;
  • 48) 0.972 042 961 043 062 784 × 2 = 1 + 0.944 085 922 086 125 568;
  • 49) 0.944 085 922 086 125 568 × 2 = 1 + 0.888 171 844 172 251 136;
  • 50) 0.888 171 844 172 251 136 × 2 = 1 + 0.776 343 688 344 502 272;
  • 51) 0.776 343 688 344 502 272 × 2 = 1 + 0.552 687 376 689 004 544;
  • 52) 0.552 687 376 689 004 544 × 2 = 1 + 0.105 374 753 378 009 088;
  • 53) 0.105 374 753 378 009 088 × 2 = 0 + 0.210 749 506 756 018 176;
  • 54) 0.210 749 506 756 018 176 × 2 = 0 + 0.421 499 013 512 036 352;

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:


0.000 000 000 742 147 478(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 478(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 478(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 00(2) × 20 =

1.1001 0111 1111 1111 1111 100(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 0111 1111 1111 1111 100


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 1011 1111 1111 1111 1100 =

100 1011 1111 1111 1111 1100


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 1011 1111 1111 1111 1100


Decimal number -0.000 000 000 742 147 478 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1100

Share

» Convert decimal -0.000 000 000 742 147 535 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 minutes

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