-0.000 000 000 742 38 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 38(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 38(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 38| = 0.000 000 000 742 38


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 38.

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 38 × 2 = 0 + 0.000 000 001 484 76;
  • 2) 0.000 000 001 484 76 × 2 = 0 + 0.000 000 002 969 52;
  • 3) 0.000 000 002 969 52 × 2 = 0 + 0.000 000 005 939 04;
  • 4) 0.000 000 005 939 04 × 2 = 0 + 0.000 000 011 878 08;
  • 5) 0.000 000 011 878 08 × 2 = 0 + 0.000 000 023 756 16;
  • 6) 0.000 000 023 756 16 × 2 = 0 + 0.000 000 047 512 32;
  • 7) 0.000 000 047 512 32 × 2 = 0 + 0.000 000 095 024 64;
  • 8) 0.000 000 095 024 64 × 2 = 0 + 0.000 000 190 049 28;
  • 9) 0.000 000 190 049 28 × 2 = 0 + 0.000 000 380 098 56;
  • 10) 0.000 000 380 098 56 × 2 = 0 + 0.000 000 760 197 12;
  • 11) 0.000 000 760 197 12 × 2 = 0 + 0.000 001 520 394 24;
  • 12) 0.000 001 520 394 24 × 2 = 0 + 0.000 003 040 788 48;
  • 13) 0.000 003 040 788 48 × 2 = 0 + 0.000 006 081 576 96;
  • 14) 0.000 006 081 576 96 × 2 = 0 + 0.000 012 163 153 92;
  • 15) 0.000 012 163 153 92 × 2 = 0 + 0.000 024 326 307 84;
  • 16) 0.000 024 326 307 84 × 2 = 0 + 0.000 048 652 615 68;
  • 17) 0.000 048 652 615 68 × 2 = 0 + 0.000 097 305 231 36;
  • 18) 0.000 097 305 231 36 × 2 = 0 + 0.000 194 610 462 72;
  • 19) 0.000 194 610 462 72 × 2 = 0 + 0.000 389 220 925 44;
  • 20) 0.000 389 220 925 44 × 2 = 0 + 0.000 778 441 850 88;
  • 21) 0.000 778 441 850 88 × 2 = 0 + 0.001 556 883 701 76;
  • 22) 0.001 556 883 701 76 × 2 = 0 + 0.003 113 767 403 52;
  • 23) 0.003 113 767 403 52 × 2 = 0 + 0.006 227 534 807 04;
  • 24) 0.006 227 534 807 04 × 2 = 0 + 0.012 455 069 614 08;
  • 25) 0.012 455 069 614 08 × 2 = 0 + 0.024 910 139 228 16;
  • 26) 0.024 910 139 228 16 × 2 = 0 + 0.049 820 278 456 32;
  • 27) 0.049 820 278 456 32 × 2 = 0 + 0.099 640 556 912 64;
  • 28) 0.099 640 556 912 64 × 2 = 0 + 0.199 281 113 825 28;
  • 29) 0.199 281 113 825 28 × 2 = 0 + 0.398 562 227 650 56;
  • 30) 0.398 562 227 650 56 × 2 = 0 + 0.797 124 455 301 12;
  • 31) 0.797 124 455 301 12 × 2 = 1 + 0.594 248 910 602 24;
  • 32) 0.594 248 910 602 24 × 2 = 1 + 0.188 497 821 204 48;
  • 33) 0.188 497 821 204 48 × 2 = 0 + 0.376 995 642 408 96;
  • 34) 0.376 995 642 408 96 × 2 = 0 + 0.753 991 284 817 92;
  • 35) 0.753 991 284 817 92 × 2 = 1 + 0.507 982 569 635 84;
  • 36) 0.507 982 569 635 84 × 2 = 1 + 0.015 965 139 271 68;
  • 37) 0.015 965 139 271 68 × 2 = 0 + 0.031 930 278 543 36;
  • 38) 0.031 930 278 543 36 × 2 = 0 + 0.063 860 557 086 72;
  • 39) 0.063 860 557 086 72 × 2 = 0 + 0.127 721 114 173 44;
  • 40) 0.127 721 114 173 44 × 2 = 0 + 0.255 442 228 346 88;
  • 41) 0.255 442 228 346 88 × 2 = 0 + 0.510 884 456 693 76;
  • 42) 0.510 884 456 693 76 × 2 = 1 + 0.021 768 913 387 52;
  • 43) 0.021 768 913 387 52 × 2 = 0 + 0.043 537 826 775 04;
  • 44) 0.043 537 826 775 04 × 2 = 0 + 0.087 075 653 550 08;
  • 45) 0.087 075 653 550 08 × 2 = 0 + 0.174 151 307 100 16;
  • 46) 0.174 151 307 100 16 × 2 = 0 + 0.348 302 614 200 32;
  • 47) 0.348 302 614 200 32 × 2 = 0 + 0.696 605 228 400 64;
  • 48) 0.696 605 228 400 64 × 2 = 1 + 0.393 210 456 801 28;
  • 49) 0.393 210 456 801 28 × 2 = 0 + 0.786 420 913 602 56;
  • 50) 0.786 420 913 602 56 × 2 = 1 + 0.572 841 827 205 12;
  • 51) 0.572 841 827 205 12 × 2 = 1 + 0.145 683 654 410 24;
  • 52) 0.145 683 654 410 24 × 2 = 0 + 0.291 367 308 820 48;
  • 53) 0.291 367 308 820 48 × 2 = 0 + 0.582 734 617 640 96;
  • 54) 0.582 734 617 640 96 × 2 = 1 + 0.165 469 235 281 92;

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 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 0001 0110 01(2)

6. Positive number before normalization:

0.000 000 000 742 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 0001 0110 01(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 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 0001 0110 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0100 0001 0110 01(2) × 20 =

1.1001 1000 0010 0000 1011 001(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 0010 0000 1011 001


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 0001 0000 0101 1001 =

100 1100 0001 0000 0101 1001


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 0001 0000 0101 1001


Decimal number -0.000 000 000 742 38 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0001 0000 0101 1001

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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