-0.000 000 000 742 147 454 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 454(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
-0.000 000 000 742 147 454(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:

|-0.000 000 000 742 147 454| = 0.000 000 000 742 147 454


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: 0.000 000 000 742 147 454.

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) 0.000 000 000 742 147 454 × 2 = 0 + 0.000 000 001 484 294 908;
  • 2) 0.000 000 001 484 294 908 × 2 = 0 + 0.000 000 002 968 589 816;
  • 3) 0.000 000 002 968 589 816 × 2 = 0 + 0.000 000 005 937 179 632;
  • 4) 0.000 000 005 937 179 632 × 2 = 0 + 0.000 000 011 874 359 264;
  • 5) 0.000 000 011 874 359 264 × 2 = 0 + 0.000 000 023 748 718 528;
  • 6) 0.000 000 023 748 718 528 × 2 = 0 + 0.000 000 047 497 437 056;
  • 7) 0.000 000 047 497 437 056 × 2 = 0 + 0.000 000 094 994 874 112;
  • 8) 0.000 000 094 994 874 112 × 2 = 0 + 0.000 000 189 989 748 224;
  • 9) 0.000 000 189 989 748 224 × 2 = 0 + 0.000 000 379 979 496 448;
  • 10) 0.000 000 379 979 496 448 × 2 = 0 + 0.000 000 759 958 992 896;
  • 11) 0.000 000 759 958 992 896 × 2 = 0 + 0.000 001 519 917 985 792;
  • 12) 0.000 001 519 917 985 792 × 2 = 0 + 0.000 003 039 835 971 584;
  • 13) 0.000 003 039 835 971 584 × 2 = 0 + 0.000 006 079 671 943 168;
  • 14) 0.000 006 079 671 943 168 × 2 = 0 + 0.000 012 159 343 886 336;
  • 15) 0.000 012 159 343 886 336 × 2 = 0 + 0.000 024 318 687 772 672;
  • 16) 0.000 024 318 687 772 672 × 2 = 0 + 0.000 048 637 375 545 344;
  • 17) 0.000 048 637 375 545 344 × 2 = 0 + 0.000 097 274 751 090 688;
  • 18) 0.000 097 274 751 090 688 × 2 = 0 + 0.000 194 549 502 181 376;
  • 19) 0.000 194 549 502 181 376 × 2 = 0 + 0.000 389 099 004 362 752;
  • 20) 0.000 389 099 004 362 752 × 2 = 0 + 0.000 778 198 008 725 504;
  • 21) 0.000 778 198 008 725 504 × 2 = 0 + 0.001 556 396 017 451 008;
  • 22) 0.001 556 396 017 451 008 × 2 = 0 + 0.003 112 792 034 902 016;
  • 23) 0.003 112 792 034 902 016 × 2 = 0 + 0.006 225 584 069 804 032;
  • 24) 0.006 225 584 069 804 032 × 2 = 0 + 0.012 451 168 139 608 064;
  • 25) 0.012 451 168 139 608 064 × 2 = 0 + 0.024 902 336 279 216 128;
  • 26) 0.024 902 336 279 216 128 × 2 = 0 + 0.049 804 672 558 432 256;
  • 27) 0.049 804 672 558 432 256 × 2 = 0 + 0.099 609 345 116 864 512;
  • 28) 0.099 609 345 116 864 512 × 2 = 0 + 0.199 218 690 233 729 024;
  • 29) 0.199 218 690 233 729 024 × 2 = 0 + 0.398 437 380 467 458 048;
  • 30) 0.398 437 380 467 458 048 × 2 = 0 + 0.796 874 760 934 916 096;
  • 31) 0.796 874 760 934 916 096 × 2 = 1 + 0.593 749 521 869 832 192;
  • 32) 0.593 749 521 869 832 192 × 2 = 1 + 0.187 499 043 739 664 384;
  • 33) 0.187 499 043 739 664 384 × 2 = 0 + 0.374 998 087 479 328 768;
  • 34) 0.374 998 087 479 328 768 × 2 = 0 + 0.749 996 174 958 657 536;
  • 35) 0.749 996 174 958 657 536 × 2 = 1 + 0.499 992 349 917 315 072;
  • 36) 0.499 992 349 917 315 072 × 2 = 0 + 0.999 984 699 834 630 144;
  • 37) 0.999 984 699 834 630 144 × 2 = 1 + 0.999 969 399 669 260 288;
  • 38) 0.999 969 399 669 260 288 × 2 = 1 + 0.999 938 799 338 520 576;
  • 39) 0.999 938 799 338 520 576 × 2 = 1 + 0.999 877 598 677 041 152;
  • 40) 0.999 877 598 677 041 152 × 2 = 1 + 0.999 755 197 354 082 304;
  • 41) 0.999 755 197 354 082 304 × 2 = 1 + 0.999 510 394 708 164 608;
  • 42) 0.999 510 394 708 164 608 × 2 = 1 + 0.999 020 789 416 329 216;
  • 43) 0.999 020 789 416 329 216 × 2 = 1 + 0.998 041 578 832 658 432;
  • 44) 0.998 041 578 832 658 432 × 2 = 1 + 0.996 083 157 665 316 864;
  • 45) 0.996 083 157 665 316 864 × 2 = 1 + 0.992 166 315 330 633 728;
  • 46) 0.992 166 315 330 633 728 × 2 = 1 + 0.984 332 630 661 267 456;
  • 47) 0.984 332 630 661 267 456 × 2 = 1 + 0.968 665 261 322 534 912;
  • 48) 0.968 665 261 322 534 912 × 2 = 1 + 0.937 330 522 645 069 824;
  • 49) 0.937 330 522 645 069 824 × 2 = 1 + 0.874 661 045 290 139 648;
  • 50) 0.874 661 045 290 139 648 × 2 = 1 + 0.749 322 090 580 279 296;
  • 51) 0.749 322 090 580 279 296 × 2 = 1 + 0.498 644 181 160 558 592;
  • 52) 0.498 644 181 160 558 592 × 2 = 0 + 0.997 288 362 321 117 184;
  • 53) 0.997 288 362 321 117 184 × 2 = 1 + 0.994 576 724 642 234 368;
  • 54) 0.994 576 724 642 234 368 × 2 = 1 + 0.989 153 449 284 468 736;

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:


0.000 000 000 742 147 454(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 454(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(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:


0.000 000 000 742 147 454(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 011


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 1011 =

100 1011 1111 1111 1111 1011


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 1011


Decimal number -0.000 000 000 742 147 454 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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