-0.000 000 000 742 147 675 45 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 45(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
-0.000 000 000 742 147 675 45(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:

|-0.000 000 000 742 147 675 45| = 0.000 000 000 742 147 675 45


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: 0.000 000 000 742 147 675 45.

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) 0.000 000 000 742 147 675 45 × 2 = 0 + 0.000 000 001 484 295 350 9;
  • 2) 0.000 000 001 484 295 350 9 × 2 = 0 + 0.000 000 002 968 590 701 8;
  • 3) 0.000 000 002 968 590 701 8 × 2 = 0 + 0.000 000 005 937 181 403 6;
  • 4) 0.000 000 005 937 181 403 6 × 2 = 0 + 0.000 000 011 874 362 807 2;
  • 5) 0.000 000 011 874 362 807 2 × 2 = 0 + 0.000 000 023 748 725 614 4;
  • 6) 0.000 000 023 748 725 614 4 × 2 = 0 + 0.000 000 047 497 451 228 8;
  • 7) 0.000 000 047 497 451 228 8 × 2 = 0 + 0.000 000 094 994 902 457 6;
  • 8) 0.000 000 094 994 902 457 6 × 2 = 0 + 0.000 000 189 989 804 915 2;
  • 9) 0.000 000 189 989 804 915 2 × 2 = 0 + 0.000 000 379 979 609 830 4;
  • 10) 0.000 000 379 979 609 830 4 × 2 = 0 + 0.000 000 759 959 219 660 8;
  • 11) 0.000 000 759 959 219 660 8 × 2 = 0 + 0.000 001 519 918 439 321 6;
  • 12) 0.000 001 519 918 439 321 6 × 2 = 0 + 0.000 003 039 836 878 643 2;
  • 13) 0.000 003 039 836 878 643 2 × 2 = 0 + 0.000 006 079 673 757 286 4;
  • 14) 0.000 006 079 673 757 286 4 × 2 = 0 + 0.000 012 159 347 514 572 8;
  • 15) 0.000 012 159 347 514 572 8 × 2 = 0 + 0.000 024 318 695 029 145 6;
  • 16) 0.000 024 318 695 029 145 6 × 2 = 0 + 0.000 048 637 390 058 291 2;
  • 17) 0.000 048 637 390 058 291 2 × 2 = 0 + 0.000 097 274 780 116 582 4;
  • 18) 0.000 097 274 780 116 582 4 × 2 = 0 + 0.000 194 549 560 233 164 8;
  • 19) 0.000 194 549 560 233 164 8 × 2 = 0 + 0.000 389 099 120 466 329 6;
  • 20) 0.000 389 099 120 466 329 6 × 2 = 0 + 0.000 778 198 240 932 659 2;
  • 21) 0.000 778 198 240 932 659 2 × 2 = 0 + 0.001 556 396 481 865 318 4;
  • 22) 0.001 556 396 481 865 318 4 × 2 = 0 + 0.003 112 792 963 730 636 8;
  • 23) 0.003 112 792 963 730 636 8 × 2 = 0 + 0.006 225 585 927 461 273 6;
  • 24) 0.006 225 585 927 461 273 6 × 2 = 0 + 0.012 451 171 854 922 547 2;
  • 25) 0.012 451 171 854 922 547 2 × 2 = 0 + 0.024 902 343 709 845 094 4;
  • 26) 0.024 902 343 709 845 094 4 × 2 = 0 + 0.049 804 687 419 690 188 8;
  • 27) 0.049 804 687 419 690 188 8 × 2 = 0 + 0.099 609 374 839 380 377 6;
  • 28) 0.099 609 374 839 380 377 6 × 2 = 0 + 0.199 218 749 678 760 755 2;
  • 29) 0.199 218 749 678 760 755 2 × 2 = 0 + 0.398 437 499 357 521 510 4;
  • 30) 0.398 437 499 357 521 510 4 × 2 = 0 + 0.796 874 998 715 043 020 8;
  • 31) 0.796 874 998 715 043 020 8 × 2 = 1 + 0.593 749 997 430 086 041 6;
  • 32) 0.593 749 997 430 086 041 6 × 2 = 1 + 0.187 499 994 860 172 083 2;
  • 33) 0.187 499 994 860 172 083 2 × 2 = 0 + 0.374 999 989 720 344 166 4;
  • 34) 0.374 999 989 720 344 166 4 × 2 = 0 + 0.749 999 979 440 688 332 8;
  • 35) 0.749 999 979 440 688 332 8 × 2 = 1 + 0.499 999 958 881 376 665 6;
  • 36) 0.499 999 958 881 376 665 6 × 2 = 0 + 0.999 999 917 762 753 331 2;
  • 37) 0.999 999 917 762 753 331 2 × 2 = 1 + 0.999 999 835 525 506 662 4;
  • 38) 0.999 999 835 525 506 662 4 × 2 = 1 + 0.999 999 671 051 013 324 8;
  • 39) 0.999 999 671 051 013 324 8 × 2 = 1 + 0.999 999 342 102 026 649 6;
  • 40) 0.999 999 342 102 026 649 6 × 2 = 1 + 0.999 998 684 204 053 299 2;
  • 41) 0.999 998 684 204 053 299 2 × 2 = 1 + 0.999 997 368 408 106 598 4;
  • 42) 0.999 997 368 408 106 598 4 × 2 = 1 + 0.999 994 736 816 213 196 8;
  • 43) 0.999 994 736 816 213 196 8 × 2 = 1 + 0.999 989 473 632 426 393 6;
  • 44) 0.999 989 473 632 426 393 6 × 2 = 1 + 0.999 978 947 264 852 787 2;
  • 45) 0.999 978 947 264 852 787 2 × 2 = 1 + 0.999 957 894 529 705 574 4;
  • 46) 0.999 957 894 529 705 574 4 × 2 = 1 + 0.999 915 789 059 411 148 8;
  • 47) 0.999 915 789 059 411 148 8 × 2 = 1 + 0.999 831 578 118 822 297 6;
  • 48) 0.999 831 578 118 822 297 6 × 2 = 1 + 0.999 663 156 237 644 595 2;
  • 49) 0.999 663 156 237 644 595 2 × 2 = 1 + 0.999 326 312 475 289 190 4;
  • 50) 0.999 326 312 475 289 190 4 × 2 = 1 + 0.998 652 624 950 578 380 8;
  • 51) 0.998 652 624 950 578 380 8 × 2 = 1 + 0.997 305 249 901 156 761 6;
  • 52) 0.997 305 249 901 156 761 6 × 2 = 1 + 0.994 610 499 802 313 523 2;
  • 53) 0.994 610 499 802 313 523 2 × 2 = 1 + 0.989 220 999 604 627 046 4;
  • 54) 0.989 220 999 604 627 046 4 × 2 = 1 + 0.978 441 999 209 254 092 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:


0.000 000 000 742 147 675 45(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 45(10) =

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


0.000 000 000 742 147 675 45(10) =

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

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

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 675 45 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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