-0.000 000 000 742 147 676 654 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 654 3(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
-0.000 000 000 742 147 676 654 3(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:

|-0.000 000 000 742 147 676 654 3| = 0.000 000 000 742 147 676 654 3


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: 0.000 000 000 742 147 676 654 3.

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) 0.000 000 000 742 147 676 654 3 × 2 = 0 + 0.000 000 001 484 295 353 308 6;
  • 2) 0.000 000 001 484 295 353 308 6 × 2 = 0 + 0.000 000 002 968 590 706 617 2;
  • 3) 0.000 000 002 968 590 706 617 2 × 2 = 0 + 0.000 000 005 937 181 413 234 4;
  • 4) 0.000 000 005 937 181 413 234 4 × 2 = 0 + 0.000 000 011 874 362 826 468 8;
  • 5) 0.000 000 011 874 362 826 468 8 × 2 = 0 + 0.000 000 023 748 725 652 937 6;
  • 6) 0.000 000 023 748 725 652 937 6 × 2 = 0 + 0.000 000 047 497 451 305 875 2;
  • 7) 0.000 000 047 497 451 305 875 2 × 2 = 0 + 0.000 000 094 994 902 611 750 4;
  • 8) 0.000 000 094 994 902 611 750 4 × 2 = 0 + 0.000 000 189 989 805 223 500 8;
  • 9) 0.000 000 189 989 805 223 500 8 × 2 = 0 + 0.000 000 379 979 610 447 001 6;
  • 10) 0.000 000 379 979 610 447 001 6 × 2 = 0 + 0.000 000 759 959 220 894 003 2;
  • 11) 0.000 000 759 959 220 894 003 2 × 2 = 0 + 0.000 001 519 918 441 788 006 4;
  • 12) 0.000 001 519 918 441 788 006 4 × 2 = 0 + 0.000 003 039 836 883 576 012 8;
  • 13) 0.000 003 039 836 883 576 012 8 × 2 = 0 + 0.000 006 079 673 767 152 025 6;
  • 14) 0.000 006 079 673 767 152 025 6 × 2 = 0 + 0.000 012 159 347 534 304 051 2;
  • 15) 0.000 012 159 347 534 304 051 2 × 2 = 0 + 0.000 024 318 695 068 608 102 4;
  • 16) 0.000 024 318 695 068 608 102 4 × 2 = 0 + 0.000 048 637 390 137 216 204 8;
  • 17) 0.000 048 637 390 137 216 204 8 × 2 = 0 + 0.000 097 274 780 274 432 409 6;
  • 18) 0.000 097 274 780 274 432 409 6 × 2 = 0 + 0.000 194 549 560 548 864 819 2;
  • 19) 0.000 194 549 560 548 864 819 2 × 2 = 0 + 0.000 389 099 121 097 729 638 4;
  • 20) 0.000 389 099 121 097 729 638 4 × 2 = 0 + 0.000 778 198 242 195 459 276 8;
  • 21) 0.000 778 198 242 195 459 276 8 × 2 = 0 + 0.001 556 396 484 390 918 553 6;
  • 22) 0.001 556 396 484 390 918 553 6 × 2 = 0 + 0.003 112 792 968 781 837 107 2;
  • 23) 0.003 112 792 968 781 837 107 2 × 2 = 0 + 0.006 225 585 937 563 674 214 4;
  • 24) 0.006 225 585 937 563 674 214 4 × 2 = 0 + 0.012 451 171 875 127 348 428 8;
  • 25) 0.012 451 171 875 127 348 428 8 × 2 = 0 + 0.024 902 343 750 254 696 857 6;
  • 26) 0.024 902 343 750 254 696 857 6 × 2 = 0 + 0.049 804 687 500 509 393 715 2;
  • 27) 0.049 804 687 500 509 393 715 2 × 2 = 0 + 0.099 609 375 001 018 787 430 4;
  • 28) 0.099 609 375 001 018 787 430 4 × 2 = 0 + 0.199 218 750 002 037 574 860 8;
  • 29) 0.199 218 750 002 037 574 860 8 × 2 = 0 + 0.398 437 500 004 075 149 721 6;
  • 30) 0.398 437 500 004 075 149 721 6 × 2 = 0 + 0.796 875 000 008 150 299 443 2;
  • 31) 0.796 875 000 008 150 299 443 2 × 2 = 1 + 0.593 750 000 016 300 598 886 4;
  • 32) 0.593 750 000 016 300 598 886 4 × 2 = 1 + 0.187 500 000 032 601 197 772 8;
  • 33) 0.187 500 000 032 601 197 772 8 × 2 = 0 + 0.375 000 000 065 202 395 545 6;
  • 34) 0.375 000 000 065 202 395 545 6 × 2 = 0 + 0.750 000 000 130 404 791 091 2;
  • 35) 0.750 000 000 130 404 791 091 2 × 2 = 1 + 0.500 000 000 260 809 582 182 4;
  • 36) 0.500 000 000 260 809 582 182 4 × 2 = 1 + 0.000 000 000 521 619 164 364 8;
  • 37) 0.000 000 000 521 619 164 364 8 × 2 = 0 + 0.000 000 001 043 238 328 729 6;
  • 38) 0.000 000 001 043 238 328 729 6 × 2 = 0 + 0.000 000 002 086 476 657 459 2;
  • 39) 0.000 000 002 086 476 657 459 2 × 2 = 0 + 0.000 000 004 172 953 314 918 4;
  • 40) 0.000 000 004 172 953 314 918 4 × 2 = 0 + 0.000 000 008 345 906 629 836 8;
  • 41) 0.000 000 008 345 906 629 836 8 × 2 = 0 + 0.000 000 016 691 813 259 673 6;
  • 42) 0.000 000 016 691 813 259 673 6 × 2 = 0 + 0.000 000 033 383 626 519 347 2;
  • 43) 0.000 000 033 383 626 519 347 2 × 2 = 0 + 0.000 000 066 767 253 038 694 4;
  • 44) 0.000 000 066 767 253 038 694 4 × 2 = 0 + 0.000 000 133 534 506 077 388 8;
  • 45) 0.000 000 133 534 506 077 388 8 × 2 = 0 + 0.000 000 267 069 012 154 777 6;
  • 46) 0.000 000 267 069 012 154 777 6 × 2 = 0 + 0.000 000 534 138 024 309 555 2;
  • 47) 0.000 000 534 138 024 309 555 2 × 2 = 0 + 0.000 001 068 276 048 619 110 4;
  • 48) 0.000 001 068 276 048 619 110 4 × 2 = 0 + 0.000 002 136 552 097 238 220 8;
  • 49) 0.000 002 136 552 097 238 220 8 × 2 = 0 + 0.000 004 273 104 194 476 441 6;
  • 50) 0.000 004 273 104 194 476 441 6 × 2 = 0 + 0.000 008 546 208 388 952 883 2;
  • 51) 0.000 008 546 208 388 952 883 2 × 2 = 0 + 0.000 017 092 416 777 905 766 4;
  • 52) 0.000 017 092 416 777 905 766 4 × 2 = 0 + 0.000 034 184 833 555 811 532 8;
  • 53) 0.000 034 184 833 555 811 532 8 × 2 = 0 + 0.000 068 369 667 111 623 065 6;
  • 54) 0.000 068 369 667 111 623 065 6 × 2 = 0 + 0.000 136 739 334 223 246 131 2;

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:


0.000 000 000 742 147 676 654 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 654 3(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:


0.000 000 000 742 147 676 654 3(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 -0.000 000 000 742 147 676 654 3 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